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A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume.
- a)24
- b)32
- c)45
- d)60
Correct answer is option 'A,C'. Can you explain this answer?
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Verified Answer
A rectangular sheet of fixed perimeter with sides having their lengths...
Let L = 8x, B = 15x and y be the length of square cut off from each corner. Then volume of box
= (8x – 2y) (15x – 2y)y
V = 120x2y – 46xy2 + 4y3
= (8x – 2y) (15x – 2y)y
V = 120x2y – 46xy2 + 4y3
Most Upvoted Answer
A rectangular sheet of fixed perimeter with sides having their lengths...
To find the maximum volume of the resulting box, we need to determine the dimensions of the box after the squares are removed. Let's consider the original rectangular sheet to have a length of 8x and a width of 15x, where x is a constant.
Finding the dimensions of the box:
- The perimeter of the original rectangular sheet is given as fixed. The perimeter P is equal to 2(length + width), which can be written as 2(8x + 15x) = 46x.
- We know that the total area of removed squares is 100. Let's assume each side of the square removed from the corners has a length of y.
- The length of the resulting box will be reduced by 2y (y from each end), and the width will be reduced by 2y as well.
- Therefore, the dimensions of the resulting box will be (8x - 2y) and (15x - 2y).
Finding the volume of the box:
- The volume V of the box is given by V = length × width × height.
- The height of the box will be y, the length of the square removed from the corners.
- Substituting the dimensions, the volume becomes V = (8x - 2y) × (15x - 2y) × y.
Finding the maximum volume:
- To find the maximum volume, we need to maximize the expression V = (8x - 2y) × (15x - 2y) × y.
- Expanding and simplifying the expression, V = 120xy - 34xy^2 + 4y^3.
- To find the maximum, we differentiate V with respect to y and equate it to zero: dV/dy = 120x - 68xy + 12y^2 = 0.
- Solving this quadratic equation for y, we get two solutions: y = 0 and y = 10x/3.
Therefore, the maximum volume occurs when y = 10x/3.
Calculating the maximum volume:
- Substituting y = 10x/3 into the volume expression, we get V = (8x - 2(10x/3)) × (15x - 2(10x/3)) × (10x/3).
- Simplifying, V = (4x/3) × (11x/3) × (10x/3) = 440x^3/27.
Comparing the answer choices:
- The maximum volume is given by V = 440x^3/27.
- We can see that the volume is directly proportional to x cubed. Therefore, the maximum volume will be maximized when x is maximized.
- Among the answer choices, the values of x that maximize the volume are 2 and 3.
- Substituting x = 2 and x = 3 into the volume expression, we get V = 1280/27 and V = 7920/27.
- The only answer choices that match these volumes are options A and C, which correspond to 24 and 45, respectively.
Therefore, the correct answer is options A and C.
Finding the dimensions of the box:
- The perimeter of the original rectangular sheet is given as fixed. The perimeter P is equal to 2(length + width), which can be written as 2(8x + 15x) = 46x.
- We know that the total area of removed squares is 100. Let's assume each side of the square removed from the corners has a length of y.
- The length of the resulting box will be reduced by 2y (y from each end), and the width will be reduced by 2y as well.
- Therefore, the dimensions of the resulting box will be (8x - 2y) and (15x - 2y).
Finding the volume of the box:
- The volume V of the box is given by V = length × width × height.
- The height of the box will be y, the length of the square removed from the corners.
- Substituting the dimensions, the volume becomes V = (8x - 2y) × (15x - 2y) × y.
Finding the maximum volume:
- To find the maximum volume, we need to maximize the expression V = (8x - 2y) × (15x - 2y) × y.
- Expanding and simplifying the expression, V = 120xy - 34xy^2 + 4y^3.
- To find the maximum, we differentiate V with respect to y and equate it to zero: dV/dy = 120x - 68xy + 12y^2 = 0.
- Solving this quadratic equation for y, we get two solutions: y = 0 and y = 10x/3.
Therefore, the maximum volume occurs when y = 10x/3.
Calculating the maximum volume:
- Substituting y = 10x/3 into the volume expression, we get V = (8x - 2(10x/3)) × (15x - 2(10x/3)) × (10x/3).
- Simplifying, V = (4x/3) × (11x/3) × (10x/3) = 440x^3/27.
Comparing the answer choices:
- The maximum volume is given by V = 440x^3/27.
- We can see that the volume is directly proportional to x cubed. Therefore, the maximum volume will be maximized when x is maximized.
- Among the answer choices, the values of x that maximize the volume are 2 and 3.
- Substituting x = 2 and x = 3 into the volume expression, we get V = 1280/27 and V = 7920/27.
- The only answer choices that match these volumes are options A and C, which correspond to 24 and 45, respectively.
Therefore, the correct answer is options A and C.
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A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume.a)24b)32c)45d)60Correct answer is option 'A,C'. Can you explain this answer?
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