Question:

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. Then the lengths of the sides of the rectangular sheet are

Updated On: Jun 14, 2022
  • 24
  • 32
  • 45
  • 60
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The Correct Option is C

Solution and Explanation



Let $l = 15\, x$ and $b = 8x$
Then Volume $= V = (8x - 2a)(15x - 2a).a$
$= 4a^3 - 46a^2x + 120ax^2$
$dV/da = 6a^2 - 46ax + 60x^2$
$d2V/da^2 = 12a - 46x$
Now, $dV/da = 0$ gives
$? 6a^2 - 46ax + 60x^2 = 0$
$? 30x^2 - 23ax + 3a^2 = 0$
$? 30x^2 - 18ax - 5ax + 3a^2 = 0$
$? 6x(5x - 3a) - a(5x - 3a) = 0$
$? (6x - a)(5x - 3a) = 0$
$? x = a/6, 3a/5$
$? x = 5/6 , 3$ when $a = 5$
When $x = 3$, $a = 5$, $d^2V/da^2 < 0$
So the Volume is maximum.
Hence, the lengths are $l = 15.3 = 45$ and $b = 8.3 = 24$.
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Concepts Used:

Application of Derivatives

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Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

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This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
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