# How do you find the cost of materials for the cheapest such container given a rectangular storage container with an open top is to have a volume of 10m^3 and the length of its base is twice the width, and the base costs $10 per square meter and material for the sides costs$6 per square meter?

Feb 7, 2015

Let $x$ and $y$ be the lenght and width of the base, and $z$ be the height.

Knowing that the lenght is twice the width means that $x = 2 y$

Knowing that the volume is fixed at $10 {m}^{3}$, means that $x \cdot y \cdot z = 10$

From the previous relation, we can express $x$ in terms of $y$, obtaining $x \cdot y \cdot z = 10 \setminus \rightarrow 2 y \cdot y \cdot z = 10 \setminus \rightarrow 2 {y}^{2} \cdot z = 10$, and from this equality we can obtain $z$ as a function of $y$: $z = \frac{10}{2 {y}^{2}} = \frac{5}{y} ^ 2$

Expressing all the three variables in terms of one is important, because now we have a problem which depends on a single variable: the area of the basis is $x \cdot y = 2 {y}^{2}$, and the lateral surface surface is given by $2 \left(x + y\right) z = 6 y \cdot \frac{5}{y} ^ 2 = \frac{30}{y}$

Since the base costs $10 per square meters, and the lateral surface costs $6 per square meters, the total cost is given by
$10 \cdot 2 {y}^{2} + 6 \cdot \frac{30}{y} = 20 {y}^{2} + \frac{180}{y}$.

So, $c \left(y\right) = 20 {y}^{2} + \frac{180}{y}$ is the function that we want to minimize, to minimize the cost. Let's study its first derivative, and find the values for which it equals zero:

$c ' \left(y\right) = 40 y - \frac{180}{y} ^ 2 = 0 \setminus \iff 40 y = \frac{180}{y} ^ 2 \setminus \iff {y}^{3} = \frac{9}{2}$
So, if you choose $y$ as the cube root of $\frac{9}{2}$, you'll minimize the function, and thus the cost.