Lagrange Multipliers: Rectangular Box

Question

Problem
Glass costs twice as much as plywood, per square meter. Use Lagrange multipliers to answer: What is the shape of the cheapest rectangular box, with 5 rectangular plywood sides and 1 rectangular glass side, that contains 1 cubic meter of volume?

Attempt
This is a question a came across which I can't wrap my head around. It doesn't look like it should be complicated yet when I try to decipher it, it reads like a riddle nearly. Anyone have any insight on how to approach this?

Answer 1

Let $a$ be the cost of plywood per square meter. Let $x$, $y$, $z$ be the dimensions of the box and let $C$ be the total cost of the box. Then we have (assuming, without loss of generality, that the glass is on a $x-y$ face) $$C=2axy+axy+ayz+ayz+axz+axz$$ that is $$C=3axy+2ayz+2axz$$ and we also know that the volume ($V$) is one cubic meter $$V=xyz=1$$

Then we have $$\nabla C= (3ay+2az,3ax+2az,2ay+2ax)$$ and $$ \nabla V=(yz,xz,yz)$$

Then we have, applying Lagrange Multiplier, there is $\lambda \neq 0$ such that $$ \nabla C = \lambda \nabla V$$ which means $$a(3y+2z,3x+2z,2y+2x)=\lambda (yz,xz,yz) $$

The solution to those equations is $x=y=2r$ and $z=3r$, where $r=\frac{2a}{\lambda}$.

Now, using that $xyz=1$, we have that $12r^3=1$ and we have that $$ \large x=y=\sqrt[3]{\frac{2}{3}} \textrm{ and } z=\sqrt[3]{\frac{9}{4}}$$

Answer 2

Let the cost of 1 square metre of plywood be 1 (the actual amount is irrelevant, we just need to set a base unit). So the cost of 1 square metre of glass is 2.

Now let the dimensions of the box be $x$, $y$, and $z$ metres, with the glass side being one of the two $x \times y$ sides. Then the cost of the box is $C(x, y, z) = 3xy + 2xz + 2yz$ (because you have two lots of $x \times z$ and $y \times z$ that cost 1 per square metre, one lot of $x \times y$ that costs 1 per square metre, and one lot of $x \times y$ that costs 2 per square metre).

Hence, you are trying to find the values of $(x, y, z)$ that minimise $C$, subject to the constraint that the volume of the box is $V(x, y, z) = xyz = 1$. Can you build your Lagrange multipler problem from there?