Optimization Problem:

A cylindrical can of height $h$ and radius $r$ is to be made. The cost of the material used for the side of the can is $C$ per square unit, and the cost of the material used for the top and bottom of the can is $2C$ per square unit. Find the dimensions of the can that minimize the cost of the material.

Note: The volume of a cylinder is given by $V = \pi r^2 h$, and the surface area is given by $S = 2\pi r^2 + 2\pi rh$.

Before we start, let's define the variables and the objective function:

• Let $r$ be the radius of the can (in units)
• Let $h$ be the height of the can (in units)
• Let $C$ be the cost per square unit for the side of the can (in cost units)
• Let $2C$ be the cost per square unit for the top and bottom of the can (in cost units)
• Let $A$ be the total cost of the material used for the can, which we want to minimize (in cost units)

The objective function we want to minimize is:

Now we can determine the expressions for the surface area of the side and the top/bottom of the can:

• The surface area of the side of the can $S_{\text{side}}$ is given by $2\pi rh$.
• The surface area of the top and bottom of the can $S_{\text{top/bottom}}$ is given by $2\pi r^2$.

Substituting these expressions into the objective function, we have:

Now we need to consider the constraint of the problem, which is the volume of the can. The volume $V$ is given by:

However, we are not directly interested in the volume. Instead, we want to express the height $h$ in terms of the radius $r$ and the volume $V$, so that we can eliminate the variable $h$ from our objective function.

Rearranging the equation for the volume, we get:

Substituting this expression for $h$ into our objective function, we have:

Simplifying, we can cancel the common factors:

Our objective now is to find the value of $r$ that minimizes the function $A(r)$.

To find the minimum of the function, we take the derivative of $A(r)$ with respect to $r$ and set it equal to zero:

Solving for $r$, we find:

Since a radius of 0 would not make sense for our context, this solution is not possible.

Hence, there is no critical point, but there is an endpoint at $r = 0$.

Therefore, we need to consider the endpoints of the possible values for $r$, as well as any critical points if they exist. In our case, we only have the endpoint $r = 0$.

Now we need to evaluate the objective function $A(r)$ at the endpoint and compare it to the values at the critical points (if they exist) to determine the minimum.

Let's evaluate the objective function at the endpoint:

Thus, the cost $A(0)$ is only $2CV$ since one of the dimensions is zero.

In conclusion, the only possible solution for minimizing the cost of the material is when the radius $r = 0$, and the height $h$ can assume any value.