# Thread: Finding the cost for the cheapest container?

1. ## Finding the cost for the cheapest container?

A rectangular container with no top has a volume of 20 m^3. The length of its base is twice the width.
The cost for the material of the base is $5 per m^2. The cost for the material of the sides are$12 per m^2.
What is the cost for the cheapest container?

My attempted work:

Let x be the width. Let y = length.
V = 20 = xyz.

C = 5xy + 12(2xz) + 12(2yz)
C = 5xy + 24xz + 24yz

V = 20 = xyz
z = 20/(xy)

C = 5xy + 240/x + 240/y

Cx = 5y - 240/(x^2)
Cy = 5x - 240/(y^2)

Not sure how to get the answer. Can somebody show me how?

2. Don't forget that the length is twice the width. That is, $y=2x$.

So:

$C = 5xy + \frac{240}{x} + \frac{240}{y}$

$\implies C = 5x(2x) + \frac{240}{x} + \frac{240}{2x}$

$\implies C = 10x^2 + \frac{240}{x} + \frac{120}{x}$

Take the derivative from here.

3. Originally Posted by drumist
Don't forget that the length is twice the width. That is, $y=2x$.

So:

$C = 5xy + \frac{240}{x} + \frac{240}{y}$
How did you get $\frac{240}{x}$

and $\frac{240}{y}$

4. Originally Posted by ione
How did you get $\frac{240}{x}$

and $\frac{240}{y}$
I copied it from what AlphaRock showed, but it looks like he made a mistake.

We have $z=\frac{20}{xy}$ and $y=2x$, so:

$C = 5xy + 24xz + 24yz$

$\implies C = 5xy + 24x\frac{20}{xy} + 24y\frac{20}{xy}$

$\implies C = 5xy + \frac{480}{y} + \frac{480}{x}$

$\implies C = 5x(2x) + \frac{480}{2x} + \frac{480}{x}$

$\implies C = 10x^2 + \frac{240}{x} + \frac{480}{x}$

$\implies C = 10x^2 + \frac{720}{x}$