Maximum volume of a box

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 _______.

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..?

• @Martín-BlasPérezPinilla They're talking about this $-1$ term inside the equation they got for $V'(h)=0$.
– Ian
Feb 20, 2021 at 20:31

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 $$62-w-h = w = h$$.

• Can you help me understand how you got to (62-w-h)h-wh = 0? (Step 2, first formula). Where did h-wh come from? Formula for volume from step 1 is 62wh - w^2h - wh^2..? Feb 21, 2021 at 8:50
• Yes I differentiated without expanding. Same thing if you expand. As you said, you have $V = 62wh - w^2h - wh^2$. Now if you differentiate with respect to $w$ and equate to zero, you get $62h - 2wh - h^2 = 0$. You divide by $h (h \ne 0)$ and you have $62 - 2w - h = 0 \implies w = 31 - \frac{h}{2}$. Feb 21, 2021 at 12:46

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.

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