# Finding the cost for the cheapest container?

• Mar 27th 2010, 09:17 PM
AlphaRock
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?
• Mar 27th 2010, 09:31 PM
drumist
Don't forget that the length is twice the width. That is, $\displaystyle y=2x$.

So:

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

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

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

Take the derivative from here.
• Mar 27th 2010, 11:12 PM
ione
Quote:

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

So:

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

How did you get $\displaystyle \frac{240}{x}$

and $\displaystyle \frac{240}{y}$
• Mar 28th 2010, 02:04 AM
drumist
Quote:

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

and $\displaystyle \frac{240}{y}$

I copied it from what AlphaRock showed, but it looks like he made a mistake.

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

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

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

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

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

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

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