Optimization

Steps to Model Optimization §

1. Draw a diagram with relevant variables (contextual domain)
2. Write a primary equation that contains the optimization variable in its isolated form, often 3+ variables
3. Write the secondary equation, which is used to reduce the primary equation down to two variables . This will never contain the optimized variable.
4. Use the secondary equation to simplify the primary equation
5. Use calculus to evaluate the extrema

Optimization Examples §

A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold $500{\text{cm}}^3$ of liquid. Find the radius and height of the can that minimizes the amount of material used.

We can use the following formulae:

$$\begin{array}{rl}{V}_{cylinder}& =\pi r^{2}h\\ A_{cylinder}& =2\pi r^2+2\pi rh\end{array}$$

First, we need to set up our primary equation. This is the equation that will contain our ‘goal’ variable. We want to find the surface area, or $A$, so our primary equation can be the normal equation for the surface area of a cylinder, as listed above.

Our secondary equation will be the one that uses the information we are given to start. We were told that the volume of the cylinder should be $500{\text{cm}}^3$, so we can just set the volume equation to that.

Our equations will end up looking like this:

$$\begin{array}{rl}\text{Primary Equation: }A& =2\pi r^2+2\pi rh\\ \text{Secondary Equation: }500& =\pi r^{2}h\end{array}$$

Next, we will want to isolate of our variables from the secondary equation, so that we can substitute it into the primary equation and basically ‘transmit’ that information over. The easiest variable to isolate would be $h$, because it is not squared in both the primary and secondary equations, and thus will be easier to work with.

To isolate $h$, we can just use simple algebra:

$$\begin{array}{rl}500& =\pi r^{2}h\\ \frac{500}{\pi r^2}& =h\end{array}$$

Then, we will substitute it into the primary equation and slightly simplify:

$$\begin{array}{rl}A& =2\pi r^2+2\pi rh\\ & \\ & =2\pi r^2+2\pi r\left(\frac{500}{\pi r^2}\right)\\ & =2\pi r^2+2\cancel{\pi r}\left(\frac{500}{\cancel{\pi }{r}^{\cancel{2}}}\right)\\ & =2\pi r^2+\frac{1000}{r}\end{array}$$

Now, we want to work on minimizing the values. To do this, we need to take the derivative of $2\pi r^2+\frac{1000}{r}$ with respect to $r$, and set it equal to zero. This is the calculus step where we find the extrema of the function. The derivative of $A$ with respect to $r$ is:

$$\begin{array}{rl}\frac{dA}{dr}& =\frac{d}{dr}\left(2\pi r^2+\frac{1000}{r}\right)\\ & =\frac{d}{dr}(2\pi r^2)+\frac{d}{dr}\left(\frac{1000}{r}\right)\\ & =4\pi r-\frac{1000}{r^2}\end{array}$$

Now we can take that and set it equal to zero to find $r$:

$$\begin{array}{rl}0& =4\pi r-\frac{1000}{r^2}\\ \frac{1000}{r^2}& =4\pi r\\ 1000& =4\pi r^3\\ \frac{1000}{4\pi }& =r^3\\ \sqrt[3]{\frac{1000}{4\pi }}& =r\\ 4.3012& \approx r\end{array}$$

Now that we have found our radius $r$, we can substitute that number into the secondary equation to derive the height $h$:

$$\begin{array}{rl}500& =\pi r^{2}h\\ & \\ \frac{500}{\pi r^2}& =h\\ \frac{500}{\pi (4.3012{)}^2}& =\\ 8.6025& \approx \end{array}$$

So to minimize our surface area, we would need to have a radius of $\approx 4.3012$ or $\sqrt[3]{\frac{1000}{4\pi }}$, and a height of $\approx 8.6025$.