Volume of open rectangular box w/o a square bottom
I am attempting to solve the following question:
An open rectangular box having a volume of 108 $\text{cm}^3$ is to be constructed from metal sheets. Find the dimensions of such a box if the amount of metal used in its construction is to be minimised.
We assume that the cost of the metal varies proportionately with the amount. Let $L$, $W$, and $H$ be the length, width, and height respectively. Let $C$ be the total cost. Since the cost varies proportionately, then $C = L + W + H$. The aim is therefore to minimise $C$.
Since $L \times W \times H = 108 \Rightarrow H = \dfrac{108}{W \times L}$.
Substituting $H$ into $C$, we obtain the equation: $C = L + W + \dfrac{108}{LW}$.
To minimise $C$, we use partial derivatives as follows:
- $\dfrac{\partial{C}}{\partial{L}} = 1 - \dfrac{108}{WL^2}$ and $\dfrac{\partial^{2}{C}}{\partial^2{L}} = \dfrac{216}{WL^3}$
- $\dfrac{\partial{C}}{\partial{L}} = 1 - \dfrac{108}{L^2W}$ and $\dfrac{\partial^{2}{C}}{\partial^2{L}} = \dfrac{216}{W^3L}$
- $\dfrac{\partial^{2}{C}}{\partial{L}\partial{W}} = \dfrac{\partial^{2}{C}}{\partial{W}\partial{L}} = \dfrac{108}{L^2 W^2}$
From $C_L = 0$, we obtain $W = \dfrac{108}{L^2}$ (*) and substituting this result into the equation $C_W = 0$, we obtain $L = \pm \sqrt{108}$. We reject $L < 0$ because the length cannot be negative.
Substituting $L = \sqrt{108}$ into (*) we obtain $W = 1$ and then substituting into $H$ we obtain $H = 1$.
Notwithstanding the proof of whether the dimensions would give the minimum, I am quite suspicious of the result of $H = 1$, $W = 1$ and $L = 108$.
Is an optimised solution even possible for this question?
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