Maximum volume of a box

Question

In his Calculus textbook (chapter 3.2, problem 23), Strang writes:

The airlines accept a box if length + width + height = $l+w+h <62''$ or 158cm. If h is fixed, show that the maximum volume $(62-w-h)wh$ is $V = h(31 - \frac{1}{2}h)^2$. Choose h to maximize V. The box with greatest volume is a _______.

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Can you help me understand the problem?

The volume is $V=whl$. Since one side is fixed, we can get that side from the formula $w+h+l=62$.

How do we get to $V = h(31 - \frac{1}{2}h)^2$?

Differentiating in terms of h: $(62-h-w)(wh) = 62-2h-1$ ? I can see that h=31 is a stationary point if disregard the -1 in my differentiation, i.e. $62-2h = 0$. I'm lost as to what to do next, and why is the -1 disregarded..?

Answer 1

As the question fixes $h$ and seeks you to find $l$ and $w$ in terms of $h$, you should differentiate wrt. $w$ and not $h$.

$V = (62-w-h) w h$

$\frac{dV}{dw} = (62-w-h) h - wh = 0 \implies 62 - 2w - h = 0$

That leads to $w = 31 - \frac{h}{2}$ and $V = h(31 - \frac{h}{2})^2$.

Now to see for which $h$ the volume is max, differentiate wrt. $h$ and equate to zero.

You should get a cube with $l = w = h$. You can confirm it by AM-GM.

$V = (62-w-h) wh$

So by AM-GM,

$\frac{(62-w-h) + w + h}{3} \geq [(62-w-h) w h]^{1/3} = V^{1/3}$

$\implies V \leq (\frac{62}{3})^3$.

Equality occurs when $62-w-h = w = h$.

Answer 2

The procedure they are describing goes like this.

• First $l+w+h=62$, since you can certainly increase the volume by lengthening any one side.
• Next, solve that for $l$: $l=62-w-h$.
• Next, fix $h$; I don't know what it is going to be yet, but regardless of what it is, once I fix $h$, $V$ is now only a function of $w$. So I look at $V(w)=(62-w-h)wh$. Since $h>0$, this is a downward-opening quadratic function, so you can find its maximum. It will happen that $w=62-w-h$ (geometrically, this is because the rectangle with fixed perimeter and largest area is a square) so $w=31-h/2$.
• Next you examine $V$ as a function of just $h$, no $w$ anymore: $V(h)=h(31-h/2)^2$, and you optimize that.

I don't know where this $62-2h-1$ came from, it is not what you get when differentiating $h(31-h/2)^2$. I think you got it by differentiating $(62-w-h)wh$ with respect to $h$, but if you do that then there are still $w$ terms floating around, so you would end up with $h$ as a function of $w$ (and you would still need to solve for $w$).

I'll also point out that there was nothing special about solving for $l$, nor about finding the optimal value of $w$ by fixing $h$. You could've dealt with the variables in any order.

Answer 3

The answer is very simple if you believe your intuition. The maximum volume occurs with a cube where each dimension is the same; call this $x$. Then $62$ inches $= 3 x$ and the maximum volume is ($62$ inches/3)^3