# Maximum volume of a cuboid box

A rectangular sheet of a 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 areas 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? (no units are specified in the question)

Now the accepted solution says sides are 8k and 15k. The side of the square removed is $$x$$. so the area of the box is $$(8k-2x)(15k-2x)x$$. Differentiate this wrt x once and then twice to find the maximum value of k possible. Substitute and find the answer.

BUT I want to ask how can we take the side of square removed to be $$x$$ when the area of 4 squares given is 100. Therefore the area of 1 square is 25. Therefore the side of the square is 5. If I have a sheet of dimensions 80cmX150cm I can remove the squares to get a box of height 5cm. If I have a sheet of length 80000kmX1500000km I can still remove a square of side 5km and make a box. Now this shows that the volume of the box is independent of height as height is fixed as 5 units. Therefore as long as the ratio is side is met, the maximum volume of the box I could make should be infinity. What am I missing?

• Perimeter is fixed, so you can't make it infinitely large Commented Jul 15, 2020 at 9:09
• Removed the tag "differential geometry". Commented Jul 15, 2020 at 9:10

Since the perimeter of the sheet is fixed, and the ratio of its sides is given to be $$8:15$$, Take the side lengths to be $$8y$$ and $$15y$$, $$\implies P = 46y$$

Say you remove a square of side $$x$$ from each corner and form a cuboid. The dimensions of the cuboid formed will be:

Length: $$15y-2x$$

Breadth: $$8y -2x$$

Height:$$\,x$$ $$\implies V = (15y - 2x)(8y - 2x)x$$ Since the perimeter P is fixed, $$y$$ must also be a constant ($$P = 46y$$)

Now to maximize Volume, just differentiate this w.r.t $$x$$ and you're done.

• my doubt is exactly about this. how are we differentiating with respect to x when x is a constant of value 5? all the values become constant then. Commented Jul 15, 2020 at 10:14
• $x$ isn't a constant! You're given that the volume is maximized when $x$ is $5$. Here, in the expression for volume, $x$ is a variable, First you need to maximize the volume, If you differentiate it and equate it to $0$, you will get $x=5$. Then you're supposed to solve for $y$ Commented Jul 15, 2020 at 10:30
• This is still not making sense to me but this being the only su, of its kind im not getting ill just copy this down. Maybe itll click later. Thanks for your help! Commented Jul 15, 2020 at 13:06
• "If the total area of the removed squares is $100$, the resulting box has maximum volume." Is the same as saying "The resulting box has maximum volume if the total area of the removed squares is $100$." When you write an expression for the volume in terms of $x$, it's a variable. You're given that it is maximum when $x=5$, so, you must first maximize the volume treating $x$ as a variable and then use the condition which says that $x=5$ for the maximum volume. Commented Jul 15, 2020 at 17:10