Cost Optimization

[racuna]

A rectangular storage container with an open top is to have a volume of 10m^3. The length of its base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

First of all, is there a formula for the volume of a rectangle?
Is it just L*W*H?

And if so, how do I optimize that?

 

[galactus]

We need to minimize surface area with the constraint \(\displaystyle xyz=10\) [1]

Let x=length

Let z=height

Let y=width

The two longest sides can be xz. Since there are two of them 2xz. Since it's $6 per square meter we have 12xz.

Same for the ends. They would be 12yz.

The base is 10xy

So we have for cost of materials with respect to surface area:

\(\displaystyle S=10xy+12xz+12yz\) [2]

Solve for z in [1].

\(\displaystyle z=\frac{10}{xy}\)

Sub into [2]

\(\displaystyle 10xy+12x(\frac{10}{xy})+12y(\frac{10}{xy})\)

\(\displaystyle S=10xy+\frac{120}{y}+\frac{120}{x}\)

Take derivative with respect to x:

\(\displaystyle \frac{\partial{S}}{\partial{x}}=10y-\frac{120}{y^{2}}\) [3]

\(\displaystyle \frac{\partial{S}}{\partial{y}}=10x-\frac{120}{y^{2}}\) [4]

Solve for y in [3]:

\(\displaystyle y=\frac{12}{x^{2}}\)

Sub into [4]

\(\displaystyle \frac{\partial{S}}{\partial{y}}=10x-\frac{120}{(\frac{12}{x^{2}})}\)

\(\displaystyle 10x-\frac{5}{6}x^{4}\)

Factor:

\(\displaystyle \frac{5x(12-x^{3})}{6}\)

As you can see, we solve for \(\displaystyle 12-x^{3}=0\) and find that

\(\displaystyle x=3^{\frac{1}{3}}\cdot2^{\frac{2}{3}}=2.29 m\)

Just sub into the equations to find y and z.

Please check my work, it's easy to make a mistake.

 

[racuna]

Thank you sooo much!
I get it now.