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Question
Solve the following:
A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares of equal area from all corners. If the total area of the removed squares is 100, the resulting box has maximum volume. Find the lengths of the rectangular sheet of paper.
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Solution
The sides of the rectangular sheet of paper are in the ratio 8 : 15.
Let the sides of the rectangular sheet of paper be 8k and 15k respectively.
Let x be the side of square which is removed from the corners of the sheet of paper.
Then total area of removed squares is 4x2, which is given to be 100.
∴ 4x2 = 100
∴ x2 = 25
∴ x = 5 ...[∵ x > 0]
Now, length, breadth and the height of the rectangular box are 15k – 2x, 8k – 2x and x respectively.
Let V be the volume of the box.
Then V = (15k – 2x)(8k – 2x).x
∴ V = (120k2 – 16kx – 30kx + 4x2).x
∴ V = 4x3 – 46kx2 + 120k2x
∴ `(dV)/dx = d/dx(4x^2 - 46k x^2 + 120k^2x)`
= 4 × 3x2 – 46k × 2x + 120k2 × 1
= 12x2 – 92kx + 120k2
Since, volume is maximum when the square of side x = 5 is removed from the corners, `((dV)/dx)_("at" x = 5)` = 0
∴ 12(5)2 – 92k(5) + 120k2 = 0
∴ 60 – 92k + 24k2 = 0
∴ 6k2 – 23k + 15 = 0
∴ 6k2 – 18k – 5k + 15 = 0
∴ 6k(k – 3) – 5(k – 3) = 0
∴ (k – 3)(6k – 5) = 0
∴ k = 3 or k = `5/6`
If k = `5/6`, then 8k – 2x = `20/3 - 10` = `(-10)/3 < 0`
∴ `k ≠ 5/6`
∴ k = 3
∴ 8k = 8 × 3 = 24 and 15k = 15 × 3 = 45
Hence, the lengths of the rectangular sheet are 24 and 45.
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