JEE Advanced 2013 · Paper 1 · Mathematics Q54

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MathematicsMultiple correct+4 / -1Question 54 of 60

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

Answer and worked solution

Official answer: A, C

Corners. $4x^2=100$ , so $x=5$ .

Volume. For sides $8k$ , $15k$ : $V=x(8k-2x)(15k-2x)$ , $V'(x)=12x^2-92kx+120k^2$ .

Maximum at x = 5: $300-460k+120k^2=0$ , i.e. $6k^2-23k+15=0$ , $k=3$ or $\frac56$ . With $k=\frac56$ the short side ( $\frac{20}3$ ) is less than $2x=10$ , impossible. So $k=3$ : sides 24 and 45.

(A), (C)

Topics:Maxima and minima

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JEE Advanced 2013 · Paper 1, Mathematics question 54