[SOLVED] optimization problem (cost function)

**coolguy00777** · May 3rd 2009, 04:27 PM

I can find the cost function, but I don't know how to find minimum cost. Which should = $329.34

A rectangular storage container with an open top is to have a volume of 20 m^3. The length of its base is twice the width. Material for the base costs $15 per square meter.Material for the sides costs $7 per square meter.
1. Find the cost function for the container.
2. Find the cost of materials for the cheapest such container.

$ V=20=2x^2y $
$ Surface Area = 2x^2+6xy $
Cost Function:
$ 15*(2x^2) + \frac{7*60}{x} $
$ 30x^2 + \frac{420}{x} $

**SengNee** · May 3rd 2009, 05:02 PM

Originally Posted by coolguy00777

I can find the cost function, but I don't know how to find minimum cost. Which should = $329.34

A rectangular storage container with an open top is to have a volume of 20 m^3. The length of its base is twice the width. Material for the base costs $15 per square meter.Material for the sides costs $7 per square meter.
1. Find the cost function for the container.
2. Find the cost of materials for the cheapest such container.

$ V=20=2x^2y $
$ Surface Area = 2x^2+6xy $
Cost Function:
$ 15*(2x^2) + \frac{7*60}{x} $
$ 30x^2 + \frac{420}{x} $

To find the max or min,you just find the stationary point of the graph and identify its concavity.

$\frac{dy}{dx}=0$
$\frac{d^2y}{dx^2}$ is positive or megative.

**coolguy00777** · May 4th 2009, 07:45 AM

sorry, but I'm still lost. Is there not an algebraic way to find the answer? What part do I take the derivative of?

**skeeter** · May 4th 2009, 07:57 AM

Originally Posted by coolguy00777

sorry, but I'm still lost. Is there not an algebraic way to find the answer? What part do I take the derivative of?

your cost function ...

$\frac{d}{dx}\left[C = 30x^2 + \frac{420}{x}\right]$

$\frac{dC}{dx} = 60x - \frac{420}{x^2} = 0$

$x = \sqrt[3]{7}$

$C(\sqrt[3]{7}) = 329.34$

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