A closed top box is to be made from piece of cardboard with a surface area of 526. If the volume is maximized, what's the length and width of the box?

Question

A closed top box with a square base is to be made from piece of cardboard with a surface area of $526$. If the volume is maximized, what's the length and width of the box?

Originally, when I tried to solve this problem, I did it by using the following formula:

Rearrange constraint equation and substitute into objective function

$V = X^2Y$

$SA = 2X^2 + 4XY = 526$

$2X^2 - 2X^2 + 4XY = 526 - 2X^2$

$4XY = 526 - 2X^2$

$4XY = 526 - 2X^2$

$\frac{4XY}{4X} = \frac{526 - 2X^2}{4X}$

$\frac{Y}{4X} = \frac{526 - 2X^2}{4X}$

$Y = 14X \cdot (526 - 2X^2)$

$V = X^2 \cdot \frac{1}{4X} \cdot (526 - 2X^2)$

$V = \frac{X^2}{4X}(526 - 2X^2)$

$V = \frac{X}{4}(526 - 2X^2)$

$V = \frac{1}{4}(526X - 2X^3)$

Next, take the derivative, set it equal to zero, and solve

$V′ = 14(526 - 6X^2) = 0$

$V′ = [14(526 - 6X^2)] 4= 0 \cdot 4$

$V′ = 526 - 526 - 6X^2 = 0 - 526$

$-6X^2 = -526$

$X = 9.36304797$

Check if the point $X = 9.36304797$ is a max, min, or saddle point.

$V′ = 1/4 \cdot (526X - 6(8)^2) = 35.5$

$V′ = 1/4 \cdot (526X - 6(10)2) = -18.5$

Derivative goes from positive, to zero, to negative, so $X = 9.36304797$ is a max.

The main problem arises when I begin to find the height (which here is done by solving for $Y$).

$Y = 1/(4X) (526 - 2X^3)$

$Y = 1/(4(9.36304797)) (526-2(9.36304797)3) = -29.78876179$

While I have found the maximum length/width, the height is a negative value (which isn't intended). Is there any issues with the formula I was using in particular? Again, I am aiming to find the maximum volume of a closed top box, which can only be achieved by finding the width/length (X) and height (Y).

Answer 1

You made an error when you factored $\frac{1}{4X}$ out of the expression $$Y = \frac{526 - 2X^2}{4X}$$ As @nickgard pointed out in the comments, it looks like you cubed $X$ when you computed $Y$ rather than squaring it, which would explain why you obtained a negative result.

Let $x$ be the width of the square base; let $h$ be the height of the box. Then \begin{align*} V & = x^2h\\ SA & = 2x^2 + 4xh \end{align*} We are given that $SA = 526$ square units. Hence, \begin{align*} 2x^2 + 4xh & = 526\\ 4xh & = 526 - 2x^2\\ h & = \frac{526 - 2x^2}{4x}\\ h & = \frac{263 - x^2}{2x} \end{align*}

Substituting the expression $\dfrac{263 - x^2}{2x}$ for $h$ in the equation for the volume gives \begin{align*} V(x) & = x^2\left(\frac{263 - x^2}{2x}\right)\\ & = \frac{1}{2}(263x - x^3) \end{align*} Differentiating yields $$V'(x) = \frac{1}{2}(263 - 3x^2)$$ Setting the derivative equal to zero yields \begin{align*} V'(x) & = 0\\ \frac{1}{2}(263 - 3x^2) & = 0\\ 263 - 3x^2 & = 0\\ -3x^2 & = 263\\ x^2 & = \frac{263}{3}\\ |x| & = \sqrt{\frac{263}{3}}\\ x & = \pm \sqrt{\frac{263}{3}} \end{align*} Since a side length cannot be negative, we conclude that the only critical point occurs at $x = \sqrt{\dfrac{263}{3}}$.

If we ignore the restriction that $x > 0$ for the moment, we notice that the graph of the function $V'(x)$ is a downward facing parabola with vertex $(0, 263/2)$ and $x$-intercepts $(-\sqrt{263/3}, 0)$ and $(\sqrt{263/3}, 0)$. Between the two $x$-intercepts, $V'(x) > 0$. Outside the two $x$-intercepts, $V'(x) < 0$. Hence, $V'(x) > 0$ if $0 < x < \sqrt{263/3}$ and $V'(x) < 0$ if $x > \sqrt{263/3}$. Hence, by the First Derivative Test, $V(x)$ has a relative maximum at $x = \sqrt{263/3}$.

Thus, the volume is maximized when $x = \sqrt{263/3}$ and \begin{align*} h & = \frac{263 - x^2}{2x}\\ & = \frac{263 - \frac{263}{3}}{2\sqrt{\frac{263}{3}}}\\ & = \frac{\frac{3 \cdot 263 - 263}{3}}{2\sqrt{\frac{263}{3}}}\\ & = \frac{2 \cdot \frac{263}{3}}{2\sqrt{\frac{263}{3}}}\\ & = \frac{\frac{263}{3}}{\sqrt{\frac{263}{3}}}\\ & = \sqrt{\frac{263}{3}} \end{align*} that is, when the box is a cube.