# Find the dimensions that will minimize the cost of the material?

##
A cylindrical container that has a capacity of #10\text( m)^3# is to be produced.

- The
*top and bottom of the container* are to be made of a material that *costs *#$20# per square meter,
- while the
*side of that container* is to be made of a material *costing *#$15# per square meter.

Find the dimensions that will minimize the cost of the material.

A cylindrical container that has a capacity of

- The
*top and bottom of the container*are to be made of a material that*costs*#$20# per square meter, - while the
*side of that container*is to be made of a material*costing*#$15# per square meter.

Find the dimensions that will minimize the cost of the material.

##### 2 Answers

#### Explanation:

Calling

Here

The volume is given by

Now, the problem can be stated as

Using lagrange multipliers it reads

with stationary points given by

solving for

We know that is a minimum because

and for the found solution has the value

For students who have not yet learned calculus of two variables, here is the single variable solution.

#### Explanation:

A (right circular) cylinder has two variables,

height,

#h# , and

radius,#r# .

The top and bottom cost

The cost of the side of the cylinder will be

The total cost is

In order to make this a function of a single variable, we need an equation with both

The volume of a (right circular) cylinder is

#h = 10/(pir^2)# .

The cost can now be written as a function of

# = 40pir^2 + 30pir(10/(pir^2))#

# = 40pir^2 + 300/r#

Domain is

We want to minimize

The only real valued critical number is

Using the first or second derivative test verifies that

Second derivative test:

So

Finally, the question asks for the dimensions, so we have

# = 10/pi (root(3)((4pi)/15))^2# (Rewrite further as desired.)