# Thread: [SOLVED] optimization problem (cost function)

1. ## [SOLVED] optimization problem (cost function)

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} $ 2. 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.

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

4. 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?
$\frac{d}{dx}\left[C = 30x^2 + \frac{420}{x}\right]$
$\frac{dC}{dx} = 60x - \frac{420}{x^2} = 0$
$x = \sqrt[3]{7}$
$C(\sqrt[3]{7}) = 329.34$