MCQ
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. The lengths of the sides of the rectangular sheet are

$(A)$ $24$ $(B)$ $32$ $(C)$ $45$ $(D)$ $60$

  • $(A,C)$
  • B
    $(B,D)$
  • C
    $(B,C)$
  • D
    $(A,D)$

Answer

Correct option: A.
$(A,C)$
a
Let $\ell=8 x, b=15 x $

$\therefore \text { Volume }=(8 x-2 a)(15 x-2 a)(a)=4 a^3-46 a^2 x+120 a x^2 $

$\frac{d V}{d a}=6 a^2-46 a x+60 x^2 $

$\left(\frac{d V}{d a}\right)_{\text {at } x=5}=0$

$\therefore \quad x=3$ and $\frac{5}{6}$

$ \frac{d^2 V}{d a^2}=6 a-23 x $

$ \left(\frac{d^2 V}{d a^2}\right)_{\text {at } a=5  x=3} < 0,$

So, at $x=3$ gives maxima

$\left(\frac{d^2 V}{d a^2}\right)_{\text {at } a=5  x=\frac{5}{6}}>0$

So, at $x=\frac{5}{6}$ gives minima.

$\frac{d V}{d a}=0 \text { when } a=5 \text { given }\left(\therefore 4 a^2=100 \text { given for maximum volume }\right) $

$\text { at } a=5 $

$\text { by } \frac{d V}{d a}=0 $

$\Rightarrow \quad 6 x^2-23 x+15=0 $

$x=3 \text { or } 5 / 6 $

So by $x=3$ (for max volume)

$8 x=24, \quad 15 x=45$

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