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 the removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are
Let the sides of squares be $x$
$$4x^2=100$$ $$x=5$$
So length of box is $l-2x$ and breadth is $b-2x$ and height is $x$
Where $\frac bl = \frac{8}{15}$
So the volume is $$V=x(b-2x)(l-2x)$$ $$V=5(b-10)(l-10)$$ $$V=5(100 -10 (l+b)+lb)$$ $$V=5(100 - 10 (\frac{23l}{15})+ \frac{8l^2}{15} )$$ Differentiating wrt $l$ $$l=\frac{230}{16}$$
The answer is $45$ and $24$ for $l$ and $b$ respectively. Where am I going wrong?
Let the side lengths be $8k$ and $15k$. After cutting a square of side $x$ from all corners, the volume of cuboid formed is: \begin{equation} V = (8k-2x)(15k-2x)(x) \end{equation}
We have to maximize it with respect to $x$ (not with respect to $k$, as you are doing). Multiplying, we get, $$V = 4x^3-46kx^2+120k^2x$$ Differentiating with respect to $x$ and equating to zero, we get, \begin{align} &12x^2-92kx+120k^2=0\\ \implies &3x^2-23kx+30k^2=0\\ \implies &x = \dfrac{23\pm13}{6}k\\ \implies &x=6k,\frac53k \end{align} But that occurs when $x=5$. Hence the values of $k$ we get are $$k=\frac56,3$$ We get the required answer for $k=3$ and $k=\frac56$ is rejected by the fact that $8k>2x$ (in order to cut from the corners).