I tried my best to come up with an answer but I failed. I would really appreciate if someone can help me with this:
Question: To carry a suitcase on an airplace, the length + width + height must be less than or equal to $ 62$ in. Assuming that height is fixed, show that the maximum volume is $ V = h(31-\frac{1}{2}h)^2 .$
Based on that, I did the following: $$l+w+h = 62$$ $$l+w = 62 - h$$
$$V = lwh$$ $$dV = dl(lwh) + dw(lwh) +dh(lwh)$$ $$dV = wh + lh + 0$$ $$dV = h(l+w)$$ $$dV = h(62-h)$$
I dont know what else to do after this...Any suggestions recommendation is greatly appreciated!
$\endgroup$ 14 Answers
$\begingroup$We want to express the volume as a function of a single variable. Since $h$ is a constant, that variable can be either $l$ or $w$. Let's use $w$.
We are given that $l + w + h = 62~\text{in}$. Hence, $l = 62~\text{in} - h - w$. Substituting $62~\text{in} - h - w$ for $l$ in the equation $V = lwh$ yields $$V(w) = (62~\text{in} - h - w)wh = (62~\text{in})wh - wh^2 - w^2h$$ Differentiating with respect to $w$, while remembering that $h$ is a constant, yields $$V'(w) = (62~\text{in})h - h^2 - 2wh$$ Setting the derivative equal to $0$ and solving for $w$ yields $$w = 31~\text{in} - \frac{h}{2}$$ Since $h > 0$, $$V''(w) = -2h < 0$$ Thus, the function has a relative maximum at $w = 31~\text{in} - \dfrac{h}{2}$ by the Second Derivative Test.
When $w = 31~\text{in} - \dfrac{h}{2}$, then $$l = 62~\text{in} - h - w = 62~\text{in} - h - \left(31~\text{in} - \frac{h}{2}\right) = 31~\text{in} - \frac{h}{2}$$ Thus, the maximum volume of the suitcase is $$V = lwh = \left(31~\text{in} - \frac{h}{2}\right)\left(31~\text{in} - \frac{h}{2}\right)h = \left(31~\text{in} - \frac{h}{2}\right)^2h$$
$\endgroup$ $\begingroup$$$l+w+h=62$$ Therefore, $L+w=62-h$. In order to maximise the volume, the area enclosed by the rectangle, $l\times w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-\frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-\frac 1 2 h)^2h$.
QED.
$\endgroup$ $\begingroup$Using Cauchy-Schwarz inequality
$ l+w+h = 62$
$l+w = 62-h$
$(l^2+w^2)(1^2+1^2) \geq (l+w)^2$
$2(l^2+w^2) \geq (l+w)^2$
$2((l+w)^2-2lw) \geq (l+w)^2$
$2(l+w)^2-4lw \geq (l+w)^2$
$(l+w)^2 \geq 4lw$
since $ lw = \frac{V}{h} $
$(62-h)^2 \geq \frac{4V}{h}$
$h(62-h)^2 \geq 4V$
$ V \leq h(31-\frac{1}{2}h)^2$
$\endgroup$ $\begingroup$Hint
To maximise the volume at fixed height you have to maximise the section of the suitcase. In this case it has to be either rectangular of square: in any case you can construct a square with the same area as a rectangle. Square section is very much easier to work with.
Solution
$\endgroup$To maximise the volume with a fixed height you have to maximise the section of the suitcase. In the simplest case, because it doesn't really matters, you can choose a square section so that $w=l$. In this case you have $$2w = 62-h\\w = 31-{1\over 2}h$$ the area of a square is $$S = w^2 = (31-{1\over 2}h)^2$$ In the end, the maximum volume is $$V=hS = h(31-{1\over 2}h)^2$$
