A cupboard with two shelves, finding the minimal coordinates to fit a fixed volume inside

Question

A cupboard with two shelves with height $y$, width $x$ and depth $d$ (in cm units).

1. Find for the required wood area to build the cupboard.
2. Find the cupboard coordinates in the least expensive way, such that the volume size inside the cupboard is $100\,\text{cm}^3$. How much would it cost if the price of the wood is $\$100$ per $100\,\text{cm}^3$?

My approach is the following: \begin{align*} &\text{Cupboard area:} &&f(x, y, d)=3xy+2yd+xd\\ &\text{Volume: } &&xyd=100. \end{align*} Now get $d$ with respect to $x$ and $y$: $$d=\frac{100}{xy},\quad f(x,y)=3xy+\frac{200}{x}+\frac{100}{y}.$$ Compute the partial derivatives and set them to zero: $$\frac{\partial f}{\partial x}=3y-\frac{200}{x^2}=0 \Longrightarrow y=\frac{200}{3x^2},\\ \frac{\partial f}{\partial y}=3x-\dfrac{100}{y^2}=0 \Longrightarrow x=\frac{100}{3y^2},\\ x= \frac{2\sqrt[3]{450}}{3},\quad y=\sqrt[3]{\frac{50}{3}},\quad d=\frac{150\sqrt[3]{3}}{\sqrt[3]{22500}}.$$

I am not sure about the answers nor the approach, I have seen almost similar problems but still not sure.

If anyone could confirm my approach or give some hints would be greatly appreciated.

Answer 1

Seems good to me, except for a few minor details:

• Seems to be a typo for y in the final answer; should be cube root, not square root. I believe the answer is $x = \sqrt[3]{\frac{400}{3}}$, $y = \sqrt[3]{\frac{50}{3}}$.
• Shouldn't it be $f(x, y, d) = 3xd + 2yd + xy$?
• You can use Lagrange multipliers, and the calculations may be a little simpler, but your method is perfectly fine.
• Technically you need to check for x = 0 or y = 0 since you assumed they are not 0 when you divided by them. But by common sense we know that the answer will not have 0 in it.

UPDATE:

Using Lagrange multipliers, let $f(x, y, d) = 3xd + 2yd + xy$, $g(x, y, d) = xyd - 100$. Then we must have $\nabla f = \lambda \nabla g$ and $g(x, y, d) = 0$. The former is $$\begin{bmatrix} 3d + y \\ 2d + x \\ 3x + 2y \end{bmatrix} = \lambda \begin{bmatrix} yd \\ xd \\ xy \\ \end{bmatrix}$$ Actually, this system of equations seems pretty hard to solve, so perhaps your method was better after all.