A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed $90 \mathrm{in}$. Suppose you want to mail a box with square sides so that its dimensions are $\mathrm{h}$ by $\mathrm{h}$ by $\mathrm{w}$ and it's girth is $2 \mathrm{~h}+2 \mathrm{w}$. What dimensions will give the box its largest volume?
A) 20 in. $x 20$ in. $x 15$ in.
B) 15 in. $x 15$ in. $x 30$ in.
C) 30 in. $x 15$ in. $x 30$ in.
D) 15 in. $x 15$ in. $x 75$ in.

Question

Quizplus · Accepted Answer

The sum of the length and girth is $2h+2w+h=3h+2w$. Since the sum cannot exceed 90, we have $3h+2w\le 90$. The volume of the box is $V=h^2w$. To maximize the volume, we need to maximize $h^2w$ subject to the constraint $3h+2w\le 90$. We can use the method of Lagrange multipliers to solve this problem. The Lagrangian is $$L(h,w,\lambda)=h^2w+\lambda(90-3h-2w)$$Setting the partial derivatives of $L$ with respect to $h$, $w$, and $\lambda$ equal to 0, we get$$\begin{align}\frac{\partial L}{\partial h}&=2hw-3\lambda=0\\frac{\partial L}{\partial w}&=h^2-2\lambda=0\\frac{\partial L}{\partial \lambda}&=90-3h-2w=0\end{align}$$From (1) and (2), we get$$\frac{h}{w}=\frac{3}{2}\Rightarrow h=\frac{3}{2}w$$Substituting this into (3), we get$$90-3\left(\frac{3}{2}w\right)-2w=0\Rightarrow w=15$$Therefore, $h=\frac{3}{2}w=22.5$. The dimensions that give the box its largest volume are 22.5 in. $x 22.5$ in. $x 15$ in.

A Private Shipping Company Will Accept a Box for Domestic 90in90 \mathrm{in}90in

A Private Shipping Company Will Accept a Box for Domestic $90 \mathrm{in}$