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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?

Aditya
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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).

Martund
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  • Isn’t $x$ a constant? Why wrt $x$? – Aditya Sep 26 '20 at 10:11
  • @Aditya, read the problem statement again. "...rectangular sheet of fixed perimeter with sides having their lengths in the ratio $\mathbf{8:15}$...If the total area of removed squares is $100$, the resulting box has maximum volume..." Note that the bold statement means the dimensions of rectangular sheet are fixed and the last statement asks us to maximize with respect to the side of corner square. – Martund Sep 26 '20 at 10:40
  • I still don’t understand. The value of $x$ is also fixed, but the dimensions aren’t, only their Addition is. – Aditya Sep 26 '20 at 12:14
  • @Aditya, The value of $x$ is not fixed. They vary $x$ and find that the volume maximizing value of $x$ is $5$. The dimensions are fixed, because both the perimeter and the side-ratio is fixed. When sum and ratio of numbers are fixed, the numbers are fixed. – Martund Sep 26 '20 at 14:19
  • Also note that they are directly saying that the perimeter and ratio of sides is fixed, whereas they say "If the total removed area is $100$, resulting cuboid has maximum volume". We can deduce that $x$ varies due to the language they use to introduce it. – Martund Sep 26 '20 at 14:24
  • That makes sense, thanks! – Aditya Sep 26 '20 at 14:50