1. ## Optimizing

I simply cannot do these problems. Putting them into some sort of workable function escapes me completely.

A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of its base is twice the width. Material for the base costs $13 per square meter. Material for the sides costs$10 per square meter. Find the cost of the materials for the cheaptest such container. Round the result to the nearest cent.

That ^ for instance. How would you even begin to approach it?

2. Originally Posted by bobsanchez
I simply cannot do these problems. Putting them into some sort of workable function escapes me completely.

A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of its base is twice the width. Material for the base costs $13 per square meter. Material for the sides costs$10 per square meter. Find the cost of the materials for the cheaptest such container. Round the result to the nearest cent.

That ^ for instance. How would you even begin to approach it?
First, assign some variables:
x = width
2x = length
h = height

The volume of this container is length x width x height = 10 m^3, or
$V = 2x \cdot x \cdot h$
$10 = 2x^2 h$

The cost of ONE side of this box would be the area of the side (in square meters) times the cost per square meter. For the base, the cost would be
$x \cdot 2x \cdot 13 = 26x^2$

Among the remaining four sides, two are longer than the other two (because the length of the base is twice the width). The cost for ONE of the longer sides is
$2x \cdot h \cdot 10 = 20xh$

The cost for ONE of the shorter sides is
$x \cdot h \cdot 10 = 10xh$

So the total cost would be the cost for the base, plus the cost of TWO longer sides, plus the cost of TWO shorter sides:
$C = 26x^2 + 20xh + 20xh + 10xh + 10xh$
$C = 26x^2 + 60xh$

We want C to be minimized, so we have two write C as a function of one variable. To do this, take the equation for V:
$10 = 2x^2 h$
and solve for h:
$h = \frac{5}{x^2}$

Plug this into C:
$C = 26x^2 + 60x \cdot \frac{5}{x^2}$
$C = 26x^2 + \frac{300}{x}$

Check the feasible domain, then set dC/dx = 0 and solve for x.

3. Okay, so I get $250.93. Is this correct? 4. That's close. It may be due to rounding, but I got about$250.90.