answer：Okay, so I have two problems here related to a modernist painter's artwork. Let me try to tackle them one by one.Starting with the first problem: The painter uses circles and squares in a ratio of 3:5. The total number of shapes is given by the polynomial (2x^2 + 7x + 3). I need to find separate polynomial expressions for the number of circles and squares.Hmm, ratios. So, if the ratio is 3:5 for circles to squares, that means for every 3 circles, there are 5 squares. So, the total number of parts in the ratio is 3 + 5 = 8 parts. Therefore, circles make up 3/8 of the total shapes, and squares make up 5/8 of the total shapes.So, if the total number of shapes is (2x^2 + 7x + 3), then the number of circles should be (frac{3}{8}(2x^2 + 7x + 3)), and the number of squares should be (frac{5}{8}(2x^2 + 7x + 3)).Let me compute that. Starting with circles:[frac{3}{8}(2x^2 + 7x + 3) = frac{3}{8} times 2x^2 + frac{3}{8} times 7x + frac{3}{8} times 3]Calculating each term:- (frac{3}{8} times 2x^2 = frac{6x^2}{8} = frac{3x^2}{4})- (frac{3}{8} times 7x = frac{21x}{8})- (frac{3}{8} times 3 = frac{9}{8})So, combining these, the number of circles is:[frac{3x^2}{4} + frac{21x}{8} + frac{9}{8}]Similarly, for squares:[frac{5}{8}(2x^2 + 7x + 3) = frac{5}{8} times 2x^2 + frac{5}{8} times 7x + frac{5}{8} times 3]Calculating each term:- (frac{5}{8} times 2x^2 = frac{10x^2}{8} = frac{5x^2}{4})- (frac{5}{8} times 7x = frac{35x}{8})- (frac{5}{8} times 3 = frac{15}{8})So, the number of squares is:[frac{5x^2}{4} + frac{35x}{8} + frac{15}{8}]Let me just double-check if these add up to the total number of shapes. Adding the number of circles and squares:[left(frac{3x^2}{4} + frac{21x}{8} + frac{9}{8}right) + left(frac{5x^2}{4} + frac{35x}{8} + frac{15}{8}right)]Combine like terms:- ( frac{3x^2}{4} + frac{5x^2}{4} = frac{8x^2}{4} = 2x^2 )- ( frac{21x}{8} + frac{35x}{8} = frac{56x}{8} = 7x )- ( frac{9}{8} + frac{15}{8} = frac{24}{8} = 3 )So, yes, it adds up to (2x^2 + 7x + 3), which matches the given total. That seems correct.Moving on to the second problem: The painter uses shades of color represented by the expression (sqrt{6x + 10} - 2). He wants the number of distinct shades to be at least 3. I need to find the range of (x) that satisfies this condition.Alright, so the expression (sqrt{6x + 10} - 2) represents the number of shades. He wants this number to be at least 3. So, the inequality is:[sqrt{6x + 10} - 2 geq 3]Let me solve this inequality step by step.First, add 2 to both sides:[sqrt{6x + 10} geq 5]Now, to eliminate the square root, I'll square both sides. But before that, I should consider the domain of the square root function. The expression inside the square root must be non-negative:[6x + 10 geq 0 implies 6x geq -10 implies x geq -frac{10}{6} = -frac{5}{3}]So, (x) must be greater than or equal to (-frac{5}{3}).Now, back to the inequality:[sqrt{6x + 10} geq 5]Squaring both sides:[6x + 10 geq 25]Subtract 10 from both sides:[6x geq 15]Divide both sides by 6:[x geq frac{15}{6} = frac{5}{2}]So, (x) must be greater than or equal to (frac{5}{2}).But wait, I should check if squaring both sides introduced any extraneous solutions. Let me test (x = frac{5}{2}):Compute the left side:[sqrt{6 times frac{5}{2} + 10} - 2 = sqrt{15 + 10} - 2 = sqrt{25} - 2 = 5 - 2 = 3]Which is equal to 3, satisfying the condition.What about (x = 3):[sqrt{6 times 3 + 10} - 2 = sqrt{18 + 10} - 2 = sqrt{28} - 2 approx 5.29 - 2 = 3.29 geq 3]Good, it works. What about (x = frac{5}{2} - 1 = frac{3}{2}):[sqrt{6 times frac{3}{2} + 10} - 2 = sqrt{9 + 10} - 2 = sqrt{19} - 2 approx 4.35 - 2 = 2.35 < 3]Which doesn't satisfy the condition. So, the solution is indeed (x geq frac{5}{2}).But wait, I also have the domain condition (x geq -frac{5}{3}). Since (frac{5}{2}) is greater than (-frac{5}{3}), the domain condition is automatically satisfied once (x geq frac{5}{2}). So, the range of (x) is (frac{5}{2}) and above.Let me write that as an interval: ([ frac{5}{2}, infty )).Wait, but the problem says "the number of distinct shades of color used to be at least 3". So, the expression (sqrt{6x + 10} - 2) must be greater than or equal to 3. So, yes, that's exactly what I solved.Just to recap:1. Start with (sqrt{6x + 10} - 2 geq 3).2. Add 2: (sqrt{6x + 10} geq 5).3. Square both sides: (6x + 10 geq 25).4. Subtract 10: (6x geq 15).5. Divide by 6: (x geq frac{5}{2}).And check the domain: (x geq -frac{5}{3}), which is less restrictive.So, the solution is (x geq frac{5}{2}).I think that's it. Let me just make sure I didn't make any calculation errors.Starting with the inequality:[sqrt{6x + 10} - 2 geq 3]Add 2:[sqrt{6x + 10} geq 5]Square:[6x + 10 geq 25]Subtract 10:[6x geq 15]Divide by 6:[x geq frac{15}{6} = frac{5}{2}]Yes, that's correct.So, summarizing:1. The number of circles is (frac{3x^2}{4} + frac{21x}{8} + frac{9}{8}), and the number of squares is (frac{5x^2}{4} + frac{35x}{8} + frac{15}{8}).2. The range of (x) is (x geq frac{5}{2}).**Final Answer**1. The number of circles is (boxed{frac{3}{4}x^2 + frac{21}{8}x + frac{9}{8}}) and the number of squares is (boxed{frac{5}{4}x^2 + frac{35}{8}x + frac{15}{8}}).2. The range of (x) is (boxed{[frac{5}{2}, infty)}).

answer：Okay, so I need to help this fire station optimize their water pressure system. There are two parts to this problem. The first one is about finding the distance where the water pressure is maximized, and the second part is figuring out the range of distances where the pressure stays above 30 psi. Let me take it step by step.Starting with the first problem: the water pressure is given by the quadratic equation ( P(d) = -0.02d^2 + 2d + 50 ). I remember that quadratic equations graph as parabolas, and since the coefficient of ( d^2 ) is negative (-0.02), the parabola opens downward. That means the vertex of the parabola is the maximum point. So, the maximum pressure occurs at the vertex.To find the vertex of a quadratic function in the form ( P(d) = ad^2 + bd + c ), the formula for the d-coordinate is ( d = -frac{b}{2a} ). Let me identify a and b from the equation. Here, a is -0.02 and b is 2.Plugging into the formula: ( d = -frac{2}{2*(-0.02)} ). Let me compute that. The denominator is 2 times -0.02, which is -0.04. So, ( d = -frac{2}{-0.04} ). Dividing 2 by 0.04 gives 50, and the negatives cancel out, so d is 50 feet. Wait, let me double-check that calculation. 2 divided by 0.04 is the same as 2 multiplied by 25, because 0.04 is 1/25. So, 2*25 is indeed 50. So, the distance at which the pressure is maximized is 50 feet.Now, to find the maximum pressure, I need to plug d = 50 back into the equation ( P(d) = -0.02d^2 + 2d + 50 ).Calculating each term:First, ( d^2 = 50^2 = 2500 ).Then, ( -0.02 * 2500 = -50 ).Next, ( 2 * 50 = 100 ).Adding the constant term, which is 50.So, putting it all together: ( -50 + 100 + 50 = 100 ). So, the maximum pressure is 100 psi.Wait a second, that seems a bit high. Let me verify the calculations again.Compute ( -0.02 * 2500 ): 0.02 * 2500 is 50, so with the negative, it's -50. Then, 2*50 is 100, and adding 50 gives -50 + 100 + 50. Hmm, that's 100. Yeah, that's correct. So, the maximum pressure is indeed 100 psi at 50 feet.Alright, moving on to the second part. The fire station wants the water pressure to remain above 30 psi. So, I need to solve the inequality ( -0.02d^2 + 2d + 50 > 30 ).First, let me subtract 30 from both sides to get the inequality in standard form:( -0.02d^2 + 2d + 50 - 30 > 0 )Simplifying that:( -0.02d^2 + 2d + 20 > 0 )So, now I have the quadratic inequality ( -0.02d^2 + 2d + 20 > 0 ). To solve this, I can first find the roots of the corresponding quadratic equation ( -0.02d^2 + 2d + 20 = 0 ), and then determine the intervals where the quadratic expression is positive.Alternatively, I can multiply both sides by -1 to make the coefficient of ( d^2 ) positive, but I have to remember that multiplying an inequality by a negative number reverses the inequality sign. Let me try that:Multiplying both sides by -1:( 0.02d^2 - 2d - 20 < 0 )So, now the inequality is ( 0.02d^2 - 2d - 20 < 0 ). Let me solve the equation ( 0.02d^2 - 2d - 20 = 0 ) to find the critical points.This is a quadratic equation, so I can use the quadratic formula:( d = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Here, a = 0.02, b = -2, c = -20.Plugging into the formula:( d = frac{-(-2) pm sqrt{(-2)^2 - 4*0.02*(-20)}}{2*0.02} )Simplify step by step:First, compute the numerator:- The first term is -(-2) which is 2.- The discriminant ( b^2 - 4ac ): ( (-2)^2 = 4 ) ( 4ac = 4*0.02*(-20) = 4*0.02*(-20) ). Let's compute 4*0.02 first, which is 0.08. Then, 0.08*(-20) is -1.6. So, the discriminant is ( 4 - (-1.6) = 4 + 1.6 = 5.6 ).So, the numerator becomes ( 2 pm sqrt{5.6} ).Compute ( sqrt{5.6} ). Let me approximate that. I know that ( sqrt{4} = 2 ) and ( sqrt{9} = 3 ), so ( sqrt{5.6} ) is somewhere between 2.3 and 2.4. Let me compute 2.36^2: 2.36*2.36. 2*2=4, 2*0.36=0.72, 0.36*2=0.72, 0.36*0.36=0.1296. Adding up: 4 + 0.72 + 0.72 + 0.1296 = 5.5696. That's very close to 5.6. So, ( sqrt{5.6} approx 2.366 ).So, the numerator is approximately ( 2 pm 2.366 ).Calculating both roots:First root: ( 2 + 2.366 = 4.366 )Second root: ( 2 - 2.366 = -0.366 )So, the roots are approximately 4.366 and -0.366.But since distance cannot be negative, we can disregard the negative root. So, the critical point is at approximately 4.366 feet.Wait, hold on. This seems odd because earlier, the maximum pressure was at 50 feet. If the quadratic crosses zero at around 4.366 feet and another negative point, but the quadratic ( 0.02d^2 - 2d - 20 ) opens upwards because the coefficient is positive. So, the quadratic is positive outside the roots and negative between them. But since one root is negative and the other is positive, the quadratic is negative between -0.366 and 4.366. But since distance can't be negative, the relevant interval where the quadratic is negative is from 0 to 4.366 feet.But wait, in our transformed inequality, we had ( 0.02d^2 - 2d - 20 < 0 ), which corresponds to the original inequality ( -0.02d^2 + 2d + 20 > 0 ). So, the solution is where the quadratic is negative, which is between the roots. But since one root is negative, the solution in the context of distance (d >= 0) is from 0 to approximately 4.366 feet.But that seems counterintuitive because the maximum pressure is at 50 feet, so the pressure should be above 30 psi from some distance beyond the hydrant up to a certain point. Wait, maybe I made a mistake in the transformation.Let me go back. The original inequality was ( -0.02d^2 + 2d + 50 > 30 ). Subtracting 30 gives ( -0.02d^2 + 2d + 20 > 0 ). Then, I multiplied both sides by -1, which reverses the inequality, so it becomes ( 0.02d^2 - 2d - 20 < 0 ). So, that's correct.But solving ( 0.02d^2 - 2d - 20 < 0 ) gives the interval between the roots where the quadratic is negative. Since one root is negative and the other is positive, the interval is from the negative root to the positive root. But since distance can't be negative, the practical interval is from 0 to approximately 4.366 feet.But that doesn't make sense because at d = 0, the pressure is 50 psi, which is above 30. At d = 50, the pressure is 100 psi, which is also above 30. So, the pressure should be above 30 psi from some distance beyond the hydrant up to a certain point. Wait, maybe I messed up the direction of the inequality when multiplying by -1.Let me think again. The original inequality after subtracting 30 is ( -0.02d^2 + 2d + 20 > 0 ). So, we can consider this quadratic ( -0.02d^2 + 2d + 20 ) and find where it's positive.Since the coefficient of ( d^2 ) is negative, the parabola opens downward. So, it will be positive between its two roots. So, the quadratic is positive between the two roots. Therefore, the solution is the interval between the two roots.But when I solved ( -0.02d^2 + 2d + 20 = 0 ), I transformed it by multiplying by -1, which gave me ( 0.02d^2 - 2d - 20 = 0 ), with roots at approximately -0.366 and 4.366. So, the original equation ( -0.02d^2 + 2d + 20 = 0 ) has roots at the same points, but since the parabola opens downward, the quadratic is positive between the two roots.But since one root is negative, the practical interval where the quadratic is positive is from 0 to 4.366 feet. But that contradicts the physical meaning because at d = 50, the pressure is 100 psi, which is above 30. So, something is wrong here.Wait, perhaps I made a mistake in solving the quadratic equation. Let me try solving ( -0.02d^2 + 2d + 20 = 0 ) directly without multiplying by -1.Using the quadratic formula:( d = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Here, a = -0.02, b = 2, c = 20.So,( d = frac{-2 pm sqrt{(2)^2 - 4*(-0.02)*(20)}}{2*(-0.02)} )Compute discriminant:( 4 - 4*(-0.02)*(20) = 4 + 0.16 = 4.16 )So,( d = frac{-2 pm sqrt{4.16}}{-0.04} )Compute ( sqrt{4.16} ). Let's see, 2^2=4, 2.04^2=4.1616, so approximately 2.04.So,First root: ( (-2 + 2.04)/(-0.04) = (0.04)/(-0.04) = -1 )Second root: ( (-2 - 2.04)/(-0.04) = (-4.04)/(-0.04) = 101 )So, the roots are at d = -1 and d = 101. Since distance can't be negative, the relevant root is at d = 101 feet.So, the quadratic ( -0.02d^2 + 2d + 20 ) is positive between d = -1 and d = 101. Since distance can't be negative, the interval where the pressure is above 30 psi is from 0 to 101 feet.Wait, that makes more sense. So, the pressure is above 30 psi from the hydrant (0 feet) up to 101 feet. Beyond 101 feet, the pressure drops below 30 psi.Let me verify this by plugging in d = 100 into the original equation:( P(100) = -0.02*(100)^2 + 2*100 + 50 = -0.02*10000 + 200 + 50 = -200 + 200 + 50 = 50 psi. That's above 30.At d = 101:( P(101) = -0.02*(101)^2 + 2*101 + 50 )Compute 101^2 = 10201So, -0.02*10201 = -204.022*101 = 202So, total pressure: -204.02 + 202 + 50 = (-204.02 + 202) + 50 = (-2.02) + 50 = 47.98 psi, which is still above 30.Wait, but according to the roots, at d = 101, the pressure should be exactly 30 psi. Hmm, perhaps my approximation was off.Wait, let me compute P(101) more accurately.( P(101) = -0.02*(101)^2 + 2*101 + 50 )101^2 = 10201-0.02*10201 = -204.022*101 = 202So, total: -204.02 + 202 + 50 = (-204.02 + 202) = -2.02 + 50 = 47.98 psi.Wait, that's still above 30. So, maybe my calculation of the roots was incorrect.Wait, let's go back to solving ( -0.02d^2 + 2d + 20 = 0 ).Using the quadratic formula:a = -0.02, b = 2, c = 20.Discriminant: ( b^2 - 4ac = 4 - 4*(-0.02)*20 = 4 + 1.6 = 5.6 )So, sqrt(5.6) ≈ 2.366Thus,d = [ -2 ± 2.366 ] / (2*(-0.02)) = [ -2 ± 2.366 ] / (-0.04)First root: (-2 + 2.366)/(-0.04) = (0.366)/(-0.04) ≈ -9.15Second root: (-2 - 2.366)/(-0.04) = (-4.366)/(-0.04) ≈ 109.15Ah, I see. Earlier, I miscalculated the denominator. It's 2a, which is 2*(-0.02) = -0.04, not -0.04 as I thought before. So, the roots are approximately -9.15 and 109.15.So, the quadratic ( -0.02d^2 + 2d + 20 ) is positive between d = -9.15 and d = 109.15. Since distance can't be negative, the interval where the pressure is above 30 psi is from 0 to approximately 109.15 feet.Let me check P(109.15):Compute ( d^2 = 109.15^2 ≈ 11914.7225 )-0.02*11914.7225 ≈ -238.294452*109.15 ≈ 218.3Adding 50: -238.29445 + 218.3 + 50 ≈ (-238.29445 + 218.3) + 50 ≈ (-19.99445) + 50 ≈ 30.00555 psi.So, approximately 30 psi at d ≈ 109.15 feet. That makes sense.Therefore, the range of distances where the pressure is above 30 psi is from 0 to approximately 109.15 feet.But since we're dealing with feet, it's practical to round to a reasonable decimal place. Let's say 109.15 feet is approximately 109.2 feet.Alternatively, we can express it as a fraction. Since 0.15 feet is about 1.8 inches, but maybe it's better to keep it in decimal form.So, summarizing:1. The maximum pressure occurs at d = 50 feet, with P = 100 psi.2. The pressure remains above 30 psi from 0 feet up to approximately 109.15 feet.Wait, but earlier when I plugged in d = 100, I got 50 psi, which is above 30. At d = 109.15, it's exactly 30 psi. So, the range is 0 < d < 109.15 feet.But let me confirm by plugging in d = 110:P(110) = -0.02*(110)^2 + 2*110 + 50 = -0.02*12100 + 220 + 50 = -242 + 220 + 50 = (-242 + 220) + 50 = (-22) + 50 = 28 psi, which is below 30. So, beyond 109.15 feet, the pressure drops below 30.Therefore, the suitable range is from 0 to 109.15 feet.Wait, but in the first part, the maximum pressure is at 50 feet, which is within this range. So, the pressure starts at 50 psi at d=0, increases to 100 psi at d=50, then decreases back to 30 psi at d≈109.15.So, the fire station needs to ensure that their hoses can handle the pressure up to 100 psi and that the effective range is up to about 109 feet.I think that's it. Let me just recap:1. Maximum pressure at d=50 feet, P=100 psi.2. Pressure above 30 psi from d=0 to d≈109.15 feet.I should present the answers clearly.

answer：Okay, so I have this problem about a bookshelf shaped like a quadratic curve. The height of the bookshelf at any point x is given by the function h(x) = -2x² + 8x + 6. There are two parts to the problem: first, finding the maximum height of the bookshelf, and second, determining where along the width of the shelf a 10-foot tall decorative piece can fit without exceeding the height of the bookshelf.Starting with the first part: finding the maximum height. Since the function is quadratic, and the coefficient of x² is negative (-2), I know the parabola opens downward, which means the vertex is the highest point. The vertex form of a quadratic function is h(x) = a(x - h)² + k, where (h, k) is the vertex. But I don't have the function in vertex form yet, so I need to find the vertex.Alternatively, I remember that for a quadratic function in standard form, h(x) = ax² + bx + c, the x-coordinate of the vertex is given by -b/(2a). Let me apply that here.Given h(x) = -2x² + 8x + 6, so a = -2, b = 8. Plugging into the formula, x = -8/(2*(-2)) = -8/(-4) = 2. So the x-coordinate of the vertex is 2 feet. To find the maximum height, I need to plug this back into the original equation.Calculating h(2): h(2) = -2*(2)² + 8*(2) + 6. Let's compute each term:-2*(4) = -88*2 = 16So, h(2) = -8 + 16 + 6 = (-8 + 16) + 6 = 8 + 6 = 14. So the maximum height is 14 feet. That seems straightforward.Wait, let me double-check my calculations. Maybe I made a mistake in arithmetic.-2*(2)^2 is -2*4 = -8. 8*2 is 16. So, -8 + 16 is 8, plus 6 is 14. Yep, that's correct. So the maximum height is 14 feet at x = 2 feet from the left end.Moving on to the second part: determining where along the width of the shelf a 10-foot tall decorative piece can fit. So, we need to find the points x where h(x) is greater than or equal to 10 feet. That is, solve the inequality -2x² + 8x + 6 ≥ 10.Let me write that inequality down:-2x² + 8x + 6 ≥ 10First, subtract 10 from both sides to bring all terms to one side:-2x² + 8x + 6 - 10 ≥ 0Simplify 6 - 10: that's -4.So, the inequality becomes:-2x² + 8x - 4 ≥ 0Hmm, maybe I can simplify this quadratic inequality. Let's see if I can factor it or perhaps divide both sides by a common factor. All coefficients are even, so let's factor out a -2 to make it easier.Factoring out -2:-2(x² - 4x + 2) ≥ 0Wait, hold on. If I factor out -2, the signs inside the parentheses will change. Let me check:-2x² + 8x - 4 = -2(x² - 4x + 2). Yes, that's correct.So, the inequality is:-2(x² - 4x + 2) ≥ 0I can divide both sides by -2, but remember that when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips.So, dividing both sides by -2:x² - 4x + 2 ≤ 0Now, I have a quadratic inequality: x² - 4x + 2 ≤ 0To solve this, I need to find the roots of the quadratic equation x² - 4x + 2 = 0, and then determine the intervals where the quadratic expression is less than or equal to zero.Let's find the roots using the quadratic formula. For ax² + bx + c = 0, the roots are x = [-b ± sqrt(b² - 4ac)]/(2a).Here, a = 1, b = -4, c = 2.So, discriminant D = b² - 4ac = (-4)^2 - 4*1*2 = 16 - 8 = 8.So, the roots are x = [4 ± sqrt(8)] / 2.Simplify sqrt(8): sqrt(8) = 2*sqrt(2). So, sqrt(8) ≈ 2.828.Therefore, the roots are:x = [4 + 2.828]/2 ≈ (6.828)/2 ≈ 3.414andx = [4 - 2.828]/2 ≈ (1.172)/2 ≈ 0.586So, the roots are approximately 0.586 and 3.414.Since the quadratic coefficient a is positive (1), the parabola opens upward. Therefore, the quadratic expression x² - 4x + 2 is less than or equal to zero between its roots.So, the solution to x² - 4x + 2 ≤ 0 is x ∈ [0.586, 3.414].But let me express this in exact terms instead of approximate decimal values. Since sqrt(8) is 2*sqrt(2), the exact roots are:x = [4 ± 2*sqrt(2)] / 2 = 2 ± sqrt(2)So, the exact roots are 2 + sqrt(2) and 2 - sqrt(2). Therefore, the solution is x ∈ [2 - sqrt(2), 2 + sqrt(2)].So, the decorative piece can be placed along the shelf from x = 2 - sqrt(2) feet to x = 2 + sqrt(2) feet.To make sure, let me verify this. Since the quadratic opens upward, the expression is negative between the roots, so the inequality x² - 4x + 2 ≤ 0 holds between 2 - sqrt(2) and 2 + sqrt(2). Therefore, the original inequality h(x) ≥ 10 holds in this interval.Let me also check a point within the interval to ensure the inequality holds. Let's pick x = 2, which is the midpoint. h(2) = 14, which is indeed greater than 10.What about x = 0? h(0) = -2*(0)^2 + 8*(0) + 6 = 6, which is less than 10. Similarly, at x = 4, h(4) = -2*(16) + 32 + 6 = -32 + 32 + 6 = 6, which is also less than 10. So, outside the interval, the height is below 10, which makes sense.Therefore, the decorative piece can be placed from x = 2 - sqrt(2) to x = 2 + sqrt(2). Calculating sqrt(2) is approximately 1.414, so 2 - 1.414 ≈ 0.586 and 2 + 1.414 ≈ 3.414, which matches my earlier approximate calculations.So, summarizing:1. The maximum height is 14 feet at x = 2 feet.2. The decorative piece can be placed from approximately 0.586 feet to 3.414 feet along the width of the shelf, or exactly from 2 - sqrt(2) to 2 + sqrt(2) feet.I think that's all. Let me just recap the steps to make sure I didn't skip anything.For the first part, identified the quadratic function, found the vertex using -b/(2a), calculated the maximum height. For the second part, set up the inequality, simplified it, solved for the roots, and determined the interval where the inequality holds. Checked with test points to ensure correctness.Yes, that seems thorough. I don't see any mistakes in my reasoning.**Final Answer**1. The maximum height of the bookshelf is boxed{14} feet.2. The decorative piece can fit along the shelf from boxed{2 - sqrt{2}} feet to boxed{2 + sqrt{2}} feet.

answer：Alright, so I have this problem about a quadratic equation modeling the number of applicants for a position over time. Let me try to figure this out step by step. First, the equation given is ( A(t) = -3t^2 + 24t + 45 ), where ( t ) is the number of months since the position was advertised. The first question is asking for the number of months after which the number of applicants reaches a maximum, and what that maximum number is. Hmm, okay. Since this is a quadratic equation, and the coefficient of ( t^2 ) is negative (-3), the parabola opens downward. That means the vertex of the parabola is the maximum point. So, I need to find the vertex of this quadratic function.I remember that for a quadratic equation in the form ( at^2 + bt + c ), the time ( t ) at which the vertex occurs is given by ( t = -frac{b}{2a} ). Let me plug in the values from the equation. Here, ( a = -3 ) and ( b = 24 ). So,( t = -frac{24}{2 times -3} )Calculating the denominator first: 2 times -3 is -6. So,( t = -frac{24}{-6} )Dividing 24 by 6 gives 4, and the negatives cancel out, so ( t = 4 ). So, the number of applicants reaches a maximum after 4 months. Now, to find the maximum number of applicants, I need to plug this value of ( t ) back into the equation ( A(t) ).Let me compute ( A(4) ):( A(4) = -3(4)^2 + 24(4) + 45 )First, calculate ( 4^2 ) which is 16. Then multiply by -3:( -3 times 16 = -48 )Next, calculate 24 times 4:( 24 times 4 = 96 )So, putting it all together:( A(4) = -48 + 96 + 45 )Adding these up: -48 + 96 is 48, and 48 + 45 is 93. Wait, that seems a bit low. Let me double-check my calculations. Wait, ( A(4) = -3(16) + 24(4) + 45 ). So, -48 + 96 is indeed 48, and 48 + 45 is 93. Hmm, okay, maybe that's correct. So, the maximum number of applicants is 93.But wait, the second part of the question is about when the number of applicants is at least 100. If the maximum is 93, then it's never reaching 100. That seems contradictory. Maybe I made a mistake in calculating the vertex.Wait, let me recalculate the vertex. The formula is ( t = -frac{b}{2a} ). Here, ( a = -3 ), ( b = 24 ). So,( t = -frac{24}{2 times -3} = -frac{24}{-6} = 4 ). So, that's correct. So, plugging back in, A(4) is indeed 93. So, the maximum number of applicants is 93, which is less than 100. So, the company never reaches 100 applicants in any month. But the second question is asking for when the number is at least 100. So, maybe I need to solve the inequality ( -3t^2 + 24t + 45 geq 100 ). Wait, but if the maximum is 93, then the equation never reaches 100. So, the inequality ( -3t^2 + 24t + 45 geq 100 ) would have no solution. Is that possible? Or maybe I made a mistake in my calculations.Let me check A(4) again:( A(4) = -3*(4)^2 + 24*4 + 45 )= -3*16 + 96 + 45= -48 + 96 + 45= (96 - 48) + 45= 48 + 45= 93Yes, that's correct. So, the maximum is 93, which is less than 100. Therefore, the inequality ( -3t^2 + 24t + 45 geq 100 ) would have no solution. But that seems odd because the question is asking to identify the range of months for which this condition is met. Maybe I need to double-check the equation or perhaps the question.Wait, let me read the problem again. It says, "the number of applicants ( A ) for a certain position each month can be modeled by the quadratic equation ( A(t) = -3t^2 + 24t + 45 ), where ( t ) is the number of months since the position was first advertised."So, the equation is correct. Then, the maximum is indeed 93, which is less than 100. Therefore, the company never reaches 100 applicants in any month. So, the inequality ( -3t^2 + 24t + 45 geq 100 ) has no solution. But the question is asking to identify the range of months for which this condition is met. Maybe I need to solve the inequality regardless.Let me proceed to solve the inequality ( -3t^2 + 24t + 45 geq 100 ).First, subtract 100 from both sides:( -3t^2 + 24t + 45 - 100 geq 0 )Simplify:( -3t^2 + 24t - 55 geq 0 )Multiply both sides by -1 to make the coefficient of ( t^2 ) positive. Remember that multiplying by a negative number reverses the inequality sign:( 3t^2 - 24t + 55 leq 0 )Now, we have a quadratic inequality ( 3t^2 - 24t + 55 leq 0 ). To find the values of ( t ) that satisfy this, we can find the roots of the quadratic equation ( 3t^2 - 24t + 55 = 0 ) and then determine the intervals where the quadratic is less than or equal to zero.Let me compute the discriminant first to see if there are real roots.Discriminant ( D = b^2 - 4ac )Here, ( a = 3 ), ( b = -24 ), ( c = 55 ).So,( D = (-24)^2 - 4*3*55 )= 576 - 660= -84Since the discriminant is negative, there are no real roots. That means the quadratic ( 3t^2 - 24t + 55 ) never crosses the t-axis and is always positive because the coefficient of ( t^2 ) is positive (3). Therefore, the inequality ( 3t^2 - 24t + 55 leq 0 ) has no solution.So, going back to the original inequality, ( -3t^2 + 24t + 45 geq 100 ) has no solution. That means there are no months where the number of applicants is at least 100. Therefore, the company never meets the target of 100 applicants in any month.But wait, the first part of the question says the maximum is 93, which is less than 100, so this makes sense. Therefore, the answer to the second question is that there is no such month where the number of applicants is at least 100.But let me confirm if I did everything correctly. Maybe I made a mistake in the inequality.Starting again, the inequality is ( -3t^2 + 24t + 45 geq 100 ).Subtract 100: ( -3t^2 + 24t - 55 geq 0 ).Multiply by -1: ( 3t^2 - 24t + 55 leq 0 ).Quadratic equation: ( 3t^2 - 24t + 55 = 0 ).Discriminant: ( (-24)^2 - 4*3*55 = 576 - 660 = -84 ). Correct.So, no real roots, quadratic is always positive, so inequality ( 3t^2 - 24t + 55 leq 0 ) has no solution.Therefore, the company never has 100 or more applicants in any month.So, summarizing:1. The number of applicants reaches a maximum after 4 months, with a maximum of 93 applicants.2. There is no month where the number of applicants is at least 100.But wait, the problem says "identify the range of months for which this condition is met." If there's no solution, maybe I should state that there are no such months. Alternatively, perhaps I made a mistake in the equation.Wait, let me check the original equation again: ( A(t) = -3t^2 + 24t + 45 ). Maybe I misread the coefficients. Let me double-check.Yes, it's -3t² +24t +45. So, that's correct.Alternatively, maybe I made a mistake in the vertex calculation. Let me recalculate A(4):( A(4) = -3*(4)^2 +24*4 +45 )= -3*16 +96 +45= -48 +96 +45= (96-48) +45= 48 +45= 93Yes, that's correct. So, the maximum is indeed 93.Therefore, the answers are:1. Maximum at 4 months, 93 applicants.2. No months where the number is at least 100.But let me think again. Maybe the company is considering the cumulative number of applicants over the months? But the problem says "the number of applicants ( A ) for a certain position each month," so it's per month, not cumulative. So, each month's applicants are modeled by that equation, and the maximum per month is 93.Therefore, the company never reaches 100 applicants in any single month.So, to answer the questions:1. After 4 months, the maximum number of applicants is 93.2. There are no months where the number of applicants is at least 100.Alternatively, if the problem expects a different interpretation, maybe considering cumulative applicants, but the problem states "the number of applicants ( A ) for a certain position each month," so it's per month, not cumulative.Therefore, I think my answers are correct.

answer：Okay, so I have this problem about a ball being thrown from a building, and I need to figure out when it will hit the ground. The equation given is ( h(t) = -16t^2 + 24t + 20 ). Hmm, I remember from school that this is a quadratic equation, and since the coefficient of ( t^2 ) is negative, the parabola opens downward. That makes sense because the ball goes up, reaches a peak, and then comes back down.Alright, the first part is to find when the ball hits the ground. That means I need to set ( h(t) = 0 ) and solve for ( t ). So, let me write that equation down:( -16t^2 + 24t + 20 = 0 )Hmm, quadratic equations can be tricky, but I think I can handle this. Maybe I can simplify it first. Let me see if I can factor out a common factor from all the terms. Looking at -16, 24, and 20, I notice they are all divisible by 4. Let me factor out a -4 to make the numbers smaller. Wait, actually, if I factor out a -4, the signs will change. Let me try factoring out a 4 instead:( 4(-4t^2 + 6t + 5) = 0 )But actually, since the original equation is equal to zero, factoring out a 4 doesn't really change anything except the coefficients. Maybe it's better to just work with the original equation. Alternatively, I could divide both sides by 4 to simplify:Dividing each term by 4:( -4t^2 + 6t + 5 = 0 )Hmm, that might be easier. So now the equation is:( -4t^2 + 6t + 5 = 0 )I could also multiply both sides by -1 to make the coefficient of ( t^2 ) positive, which might make it easier to factor or apply the quadratic formula. Let me do that:( 4t^2 - 6t - 5 = 0 )Okay, now it's ( 4t^2 - 6t - 5 = 0 ). Let me see if this can be factored. I need two numbers that multiply to ( 4 * (-5) = -20 ) and add up to -6. Hmm, let's think: factors of -20 could be -10 and 2, which add up to -8. Not quite. How about -5 and 4? That adds up to -1. Not it. Maybe -8 and 2.5? Wait, that's not integers. Maybe this doesn't factor nicely. If that's the case, I should use the quadratic formula.Quadratic formula is ( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). In this equation, ( a = 4 ), ( b = -6 ), and ( c = -5 ). Plugging those in:First, compute the discriminant ( b^2 - 4ac ):( (-6)^2 - 4*4*(-5) = 36 - (-80) = 36 + 80 = 116 )So, the discriminant is 116. That's a positive number, so we have two real solutions. Now, compute the roots:( t = frac{-(-6) pm sqrt{116}}{2*4} = frac{6 pm sqrt{116}}{8} )Simplify ( sqrt{116} ). Let's see, 116 can be broken down into 4*29, so ( sqrt{116} = sqrt{4*29} = 2sqrt{29} ). So, substituting back:( t = frac{6 pm 2sqrt{29}}{8} )We can factor out a 2 in the numerator:( t = frac{2(3 pm sqrt{29})}{8} = frac{3 pm sqrt{29}}{4} )So, the two solutions are:( t = frac{3 + sqrt{29}}{4} ) and ( t = frac{3 - sqrt{29}}{4} )Now, let's compute these numerically to see which one makes sense in the context of the problem.First, ( sqrt{29} ) is approximately 5.385. So,( t = frac{3 + 5.385}{4} = frac{8.385}{4} approx 2.096 ) seconds.And,( t = frac{3 - 5.385}{4} = frac{-2.385}{4} approx -0.596 ) seconds.Since time cannot be negative, we discard the negative solution. Therefore, the ball will hit the ground at approximately 2.096 seconds.Wait a second, let me verify my calculations because 2.096 seems a bit short for a ball thrown from a 20-foot building with an initial velocity of 24 feet per second. Maybe I made a mistake in simplifying the equation earlier.Let me go back to the original equation:( -16t^2 + 24t + 20 = 0 )Instead of dividing by 4, maybe I should apply the quadratic formula directly here. Let me try that.Here, ( a = -16 ), ( b = 24 ), ( c = 20 ).Discriminant ( D = b^2 - 4ac = 24^2 - 4*(-16)*20 = 576 + 1280 = 1856 )Wait, that's different from before. Earlier, after dividing by 4, I had discriminant 116, but now it's 1856. That makes sense because 1856 divided by 16 (since we divided by 4 earlier, which squared is 16) is 116. So, that's consistent.So, using the original coefficients:( t = frac{-24 pm sqrt{1856}}{2*(-16)} )Simplify ( sqrt{1856} ). Let's see, 1856 divided by 16 is 116, so ( sqrt{1856} = sqrt{16*116} = 4sqrt{116} ). And earlier, we saw that ( sqrt{116} = 2sqrt{29} ), so ( sqrt{1856} = 4*2sqrt{29} = 8sqrt{29} ).So, substituting back:( t = frac{-24 pm 8sqrt{29}}{-32} )We can factor out a -8 from numerator and denominator:( t = frac{-8(3 mp sqrt{29})}{-32} = frac{3 mp sqrt{29}}{4} )So, the solutions are:( t = frac{3 - sqrt{29}}{4} ) and ( t = frac{3 + sqrt{29}}{4} )Wait, that's the same as before, just written differently. So, plugging in the approximate value:( t approx frac{3 + 5.385}{4} approx 2.096 ) seconds and ( t approx frac{3 - 5.385}{4} approx -0.596 ) seconds.So, same result. So, the positive solution is approximately 2.096 seconds. But as I thought earlier, that seems a bit short. Let me check if my initial equation is correct.The equation given is ( h(t) = -16t^2 + 24t + 20 ). The standard form for projectile motion is ( h(t) = -16t^2 + v_0 t + h_0 ), where ( v_0 ) is the initial velocity and ( h_0 ) is the initial height. So, in this case, ( v_0 = 24 ) ft/s and ( h_0 = 20 ) ft. That seems correct.Let me plug t = 2.096 into the equation to see if h(t) is approximately 0.Compute ( h(2.096) = -16*(2.096)^2 + 24*(2.096) + 20 )First, compute ( (2.096)^2 approx 4.394 )So, ( -16*4.394 approx -70.304 )Then, ( 24*2.096 approx 50.304 )Add them up: -70.304 + 50.304 + 20 = (-70.304 + 50.304) + 20 = (-20) + 20 = 0. Perfect, so that checks out.But intuitively, 2 seconds seems quick for a 20-foot building. Let me think about the initial velocity. 24 feet per second is roughly 16 miles per hour, which is pretty fast for a thrown ball. Maybe it's reasonable.Alternatively, let me compute the time it takes to reach the maximum height and see if that adds up.The time to reach the vertex (maximum height) is at ( t = -b/(2a) ). In the original equation, ( a = -16 ), ( b = 24 ).So, ( t = -24/(2*(-16)) = -24/(-32) = 0.75 ) seconds. So, the ball reaches its peak at 0.75 seconds. Then, it takes another approximately 1.346 seconds to come down (2.096 - 0.75 = 1.346). That seems plausible.Alternatively, let me compute the maximum height to see if that makes sense.At t = 0.75, ( h(0.75) = -16*(0.75)^2 + 24*(0.75) + 20 )Compute ( (0.75)^2 = 0.5625 )So, ( -16*0.5625 = -9 )( 24*0.75 = 18 )So, ( h(0.75) = -9 + 18 + 20 = 29 ) feet. So, the ball goes up to 29 feet, which is 9 feet above the building. That seems reasonable given the initial velocity.So, from 29 feet, it falls back down to the ground. The time to fall from 29 feet can be calculated using the equation ( h = frac{1}{2}gt^2 ), but since we already have the quadratic equation, it's accounted for in the total time.So, all in all, 2.096 seconds seems correct. Maybe it's quicker than I thought because of the initial upward velocity.Now, moving on to the second part: verifying if the solution is realistic by considering the initial and final conditions.At t = 0, the height should be 20 feet, which is the height of the building. Plugging t = 0 into the equation:( h(0) = -16*(0)^2 + 24*(0) + 20 = 20 ) feet. That's correct.At t = 2.096, we already saw that h(t) = 0, which is the ground. So, that's consistent.Additionally, the ball reaches a maximum height of 29 feet at t = 0.75 seconds, which is higher than the building, so that makes sense with the initial upward velocity.Therefore, the solution t ≈ 2.096 seconds is realistic.Wait, but let me think again. If the ball is thrown upward, it goes up, slows down, stops at the peak, and then starts falling. The total time should be the time to go up plus the time to come down. Since the time to go up is 0.75 seconds, the time to come down should be longer than that because the ball has to cover the 29 feet back down, but also, it's starting from rest at the peak.Wait, actually, no. The time to go up and the time to come down aren't necessarily the same because the initial velocity is upward, but the acceleration is downward. However, in projectile motion without air resistance, the time to reach the peak is equal to the time to fall back to the same height if it were starting from the peak. But in this case, the ball isn't starting from the peak; it's starting from the building.Wait, actually, the time to go up is 0.75 seconds, and the time to come down from the peak to the ground is another 1.346 seconds, which is longer than the time to go up. That makes sense because the ball has to cover more distance on the way down (from 29 feet to 0) than on the way up (from 20 feet to 29 feet). So, the times aren't equal, which is fine.Alternatively, if the ball were thrown from the ground, the time up and time down would be equal, but since it's thrown from a height, the time down is longer.So, all in all, the solution seems consistent and realistic.

answer：Okay, so I have this problem about Mr. Norman Gould Allen and his friend planting a community garden. The total area is given by the polynomial ( A(x) = x^3 - 6x^2 + 11x - 6 ). I need to find the dimensions of the sections by finding the roots of this polynomial. Then, for the vegetable section, which is three times longer than it is wide, I have to write and solve a quadratic inequality to find the range of possible widths that keep the area less than 30 square meters.Alright, let's start with the first part: finding the roots of the polynomial ( A(x) = x^3 - 6x^2 + 11x - 6 ). Since it's a cubic polynomial, I know it can have up to three real roots. To find the roots, I can try factoring it. Maybe I can factor by grouping or use the Rational Root Theorem.The Rational Root Theorem says that any possible rational root, p/q, is such that p is a factor of the constant term and q is a factor of the leading coefficient. Here, the constant term is -6, and the leading coefficient is 1. So the possible rational roots are ±1, ±2, ±3, ±6.Let me test these one by one. I'll start with x=1:( A(1) = 1 - 6 + 11 - 6 = 0 ). Oh, so x=1 is a root. That means (x - 1) is a factor.Now, I can perform polynomial division or use synthetic division to factor out (x - 1) from the polynomial. Let's do synthetic division.Set up the coefficients: 1 (for x³), -6 (for x²), 11 (for x), -6 (constant term).Bring down the 1. Multiply by 1: 1*1=1. Add to next coefficient: -6 +1 = -5. Multiply by 1: -5*1=-5. Add to next coefficient: 11 + (-5)=6. Multiply by 1: 6*1=6. Add to last coefficient: -6 +6=0. Perfect, no remainder.So after factoring out (x - 1), the polynomial becomes (x - 1)(x² - 5x + 6). Now, let's factor the quadratic: x² - 5x + 6.Looking for two numbers that multiply to 6 and add to -5. Those would be -2 and -3.So, x² - 5x + 6 factors into (x - 2)(x - 3). Therefore, the full factorization of A(x) is (x - 1)(x - 2)(x - 3).So the roots are x=1, x=2, and x=3. These are the potential dimensions of each section. I think that answers the first part.Now, moving on to the second part. The vegetable section is three times longer than it is wide, and the width is represented by one of the roots. So the width is either 1, 2, or 3 meters, and the length is three times that.But wait, the vegetable section is a part of the garden, so its dimensions must be among the roots. Hmm, but the roots are 1, 2, 3. So if the width is one of these, the length would be 3, 6, or 9 meters. But the total area is given by the polynomial, which is a cubic, so maybe the sections are two-dimensional? Or perhaps the garden is divided into two parts, each with dimensions based on the roots.Wait, the problem says the total area is represented by the polynomial, and the factors represent the potential dimensions. So maybe the garden is a three-dimensional space? Or perhaps it's a two-dimensional area, but the polynomial is a cubic, which is a bit confusing.Wait, maybe the garden is divided into two sections, each with an area, and the total area is the product of their areas? Or maybe the garden is a three-dimensional space, but that seems less likely for a garden. Hmm.Wait, the problem says the total area is represented by the polynomial, so it's a two-dimensional area, but the polynomial is cubic. That suggests that the garden might be divided into three sections, but the problem says two sections: one for vegetables and one for flowers. So perhaps each section is a rectangle with dimensions given by two of the roots, and the total area is the product of the two areas? Or maybe the total area is the sum of the two areas?Wait, no, the polynomial is given as ( A(x) = x^3 - 6x^2 + 11x - 6 ). So if x is a variable related to the dimensions, maybe the garden is a three-dimensional space? But that doesn't make much sense for a garden. Alternatively, perhaps the garden is divided into two sections, each with an area that is a quadratic function, and the total area is the product of these two quadratics? But then the total area would be a quartic, not a cubic.Hmm, perhaps I need to think differently. Maybe the garden is divided into two sections, each with dimensions given by two of the roots, and the total area is the sum of their areas. Let me check.Wait, if the garden is divided into two sections, each with area given by a quadratic, then the total area would be the sum of two quadratics, which would be a quadratic, not a cubic. So that doesn't fit.Alternatively, maybe the garden is a three-dimensional space, with length, width, and height, but that seems less likely. Alternatively, perhaps the garden is divided into two sections, each with a certain dimension, and the total area is a cubic function because it's a product of three dimensions? But the problem mentions two sections, so maybe each section has two dimensions, and the total area is the product of the two areas? That would make sense.Wait, let's think. If the garden is divided into two sections, each with area, say, A1 and A2, then the total area would be A1 + A2. But the problem says the total area is given by the polynomial, which is a cubic. So A1 + A2 = x³ -6x² +11x -6.But if each section is a rectangle, then A1 = length1 * width1, and A2 = length2 * width2. If the sections are divided such that one dimension is shared, maybe? For example, if the garden is a larger rectangle, divided into two smaller rectangles along one dimension, then the total area would be the sum of the two smaller areas.But in that case, the total area would be (length1 + length2) * width, which is a quadratic, not a cubic. Hmm.Wait, maybe the garden is a three-dimensional space, with length, width, and height, but that seems odd. Alternatively, perhaps the garden is divided into two sections, each of which is a three-dimensional space, but again, that seems less likely.Wait, perhaps the polynomial represents the area as a function of x, where x is a dimension, and the garden is divided into two sections, each with dimensions related to x. For example, maybe the garden is a rectangle with length x and width something, and it's divided into two sections with areas given by two quadratics, whose sum is the cubic polynomial.But that might be overcomplicating it. Alternatively, perhaps the garden is divided into two sections, each with an area that is a quadratic function of x, and the total area is the product of these two quadratics, which would be a quartic, but the given polynomial is cubic, so that doesn't fit.Wait, maybe the garden is divided into two sections, each with a linear dimension, and the total area is the product of these two linear dimensions and another dimension. Hmm, but that would make the total area a cubic, which matches the given polynomial.So, if the garden is a three-dimensional space, with length, width, and height, then the total volume would be a cubic polynomial. But the problem mentions area, not volume. So that might not be it.Wait, perhaps the garden is a two-dimensional area, but the polynomial is cubic because it's a function of x, which could represent a dimension, and the area is expressed in terms of x. For example, if the garden is a rectangle with length x and width something, then the area would be quadratic. But here it's cubic, so maybe it's a more complex shape.Alternatively, perhaps the garden is divided into two sections, each with an area that is a quadratic function of x, and the total area is the sum, which is a cubic. But that would require each section to have an area that is quadratic, which might be possible.Wait, maybe the garden is divided into two sections, each of which is a rectangle with one side x and the other side being a linear function of x. So, for example, one section has area x*(ax + b) and the other has area x*(cx + d), and their sum is x*(ax + b + cx + d) = x*( (a + c)x + (b + d) ), which is quadratic, not cubic. So that doesn't fit.Hmm, perhaps the garden is divided into two sections, each of which is a rectangle with sides that are linear functions of x, so their areas would be quadratic, and the total area would be the sum, which is quadratic. But the given polynomial is cubic, so that doesn't fit.Wait, maybe the garden is divided into two sections, each of which is a three-dimensional space, but again, that seems odd.Alternatively, perhaps the garden is a three-dimensional space, and the total volume is given by the cubic polynomial, but the problem mentions area, so that's confusing.Wait, maybe the garden is divided into two sections, each with an area that is a quadratic function, and the total area is the product of these two quadratics, which would be a quartic, but the given polynomial is cubic, so that doesn't fit.Hmm, I'm getting stuck here. Maybe I need to approach it differently. The problem says the factors of the polynomial represent the potential dimensions of each section. So the roots are 1, 2, 3. So the dimensions could be 1, 2, 3 meters. Since it's a garden divided into two sections, maybe each section has two dimensions, and the total area is the product of these two areas, which would be a quartic, but the given polynomial is cubic. Hmm.Wait, perhaps the garden is a three-dimensional space, with length, width, and height, and the total volume is given by the cubic polynomial. Then, the two sections would each have their own dimensions, and the total volume is the sum of their volumes. But the problem mentions area, so that's conflicting.Wait, maybe the garden is divided into two sections, each with an area that is a quadratic function, and the total area is the product of these two quadratics, which would be a quartic, but the given polynomial is cubic. So that doesn't fit.Alternatively, perhaps the garden is divided into two sections, each with an area that is a linear function, and the total area is the product, which would be quadratic, but the given polynomial is cubic.Hmm, I'm not sure. Maybe I need to think of the garden as a three-dimensional space, with length, width, and height, and the total volume is given by the cubic polynomial. Then, each section would have its own dimensions, and the total volume is the sum of their volumes. But the problem mentions area, so that's conflicting.Wait, maybe the garden is divided into two sections, each with an area that is a quadratic function, and the total area is the sum, which is a cubic. So, for example, if one section has area ( x^2 ) and the other has area ( -6x^2 + 11x -6 ), but that seems arbitrary.Alternatively, perhaps the garden is divided into two sections, each with an area that is a linear function, and the total area is the product, which is quadratic, but the given polynomial is cubic.Wait, maybe the garden is divided into two sections, each with an area that is a cubic function, but that seems too complex.Wait, perhaps the garden is divided into two sections, each with an area that is a quadratic function, and the total area is the product, which would be a quartic, but the given polynomial is cubic.Hmm, I'm going in circles here. Maybe I should focus on the first part, which I think I've solved: the roots are 1, 2, 3. So the dimensions are 1, 2, 3 meters.Now, moving on to the second part. The vegetable section is three times longer than it is wide, and the width is represented by one of the roots. So the width is either 1, 2, or 3 meters, and the length is three times that, so 3, 6, or 9 meters.But wait, the area of the vegetable section would be width * length, which is width * 3*width = 3*width². We need this area to be less than 30 square meters.So, 3*width² < 30. Let's write that as an inequality:3w² < 30Divide both sides by 3:w² < 10Take square roots:w < √10 ≈ 3.162But the width is one of the roots, which are 1, 2, 3. So we need to find which of these satisfy w < √10.√10 is approximately 3.162, so w can be 1, 2, or 3, since all are less than 3.162. Wait, but 3 is less than 3.162, so all three are valid.But wait, the problem says to write and solve a quadratic inequality. So maybe I need to consider the width as a variable, not just the roots. Wait, the problem says the width is represented by one of the roots, so it's fixed. But the inequality is about the area being less than 30, so maybe the width can vary, but it's constrained by the roots.Wait, perhaps I misinterpreted. Maybe the width is a variable, and the length is three times the width, and the area of the vegetable section is 3w², which needs to be less than 30. So the inequality is 3w² < 30, which simplifies to w² < 10, so w < √10. But since the width must be one of the roots, which are 1, 2, 3, then all of them satisfy w < √10, so all are possible.But that seems too straightforward. Maybe I need to consider that the width is a variable, not necessarily one of the roots, but related to the roots. Wait, the problem says the width is represented by one of the roots, so it's fixed. So the width is either 1, 2, or 3, and the length is 3, 6, or 9, respectively. Then, the area would be 3, 12, or 27 square meters, all of which are less than 30. So all three are possible.But the problem says to write and solve a quadratic inequality. So maybe I need to consider the width as a variable, not fixed to the roots, but constrained by the roots. Wait, perhaps the width is a variable, and the length is three times the width, and the area is 3w², which must be less than 30. So the inequality is 3w² < 30, which simplifies to w² < 10, so w < √10. But since the width is represented by one of the roots, which are 1, 2, 3, then the possible widths are 1, 2, 3, all less than √10, so all are valid.But that seems too simple. Maybe I need to consider that the width is a variable, and the length is three times the width, and the area is 3w², which must be less than 30. So the inequality is 3w² < 30, which simplifies to w² < 10, so w < √10. Since the width must be a positive real number, the solution is 0 < w < √10.But the problem mentions that the width is represented by one of the roots, which are 1, 2, 3. So the possible widths are 1, 2, 3, all of which are less than √10, so all are valid. Therefore, the range of possible widths is w ∈ {1, 2, 3}.But the problem says to write and solve a quadratic inequality, so maybe I need to express it in terms of w, not just state the roots. So let's write the inequality:3w² < 30Divide both sides by 3:w² < 10So the solution is w < √10 and w > -√10. But since width can't be negative, we have 0 < w < √10.But since the width is one of the roots, which are positive integers, the possible widths are 1, 2, 3.Therefore, the range of possible widths is 1, 2, 3 meters.Wait, but the problem says to write and solve a quadratic inequality, so maybe I need to present it as such, even though the roots are specific. So the inequality is w² < 10, which gives w ∈ (-√10, √10), but since width is positive, w ∈ (0, √10). But since the width is one of the roots, which are 1, 2, 3, all within this interval, so all are acceptable.So, to summarize:1. The roots of the polynomial are 1, 2, 3, so the dimensions are 1m, 2m, 3m.2. The vegetable section has width w (one of the roots) and length 3w. The area is 3w² < 30, which simplifies to w² < 10, so w < √10. Since w is 1, 2, or 3, all satisfy this, so the possible widths are 1, 2, 3 meters.I think that's it.