# Optimization

Optimization refers to the maximums or minimums of a function in calculus.

Common types of optimization problems include:

Optimization Area & Perimeter

Optimization Volume & Surface Area

Optimization of the Distance Between a Point on a Curve

First step is to create two or more equations to solve the problem.

Second step is to find the global maximum or minimum through the first derivative test.

Third step is to substitute the variable back into one of the equations to find the other variables.

## Example 1

A rectangle has its base on the x-axis and its upper two vertices on the parabola \(y = 3 − x^2\), as shown in the figure below. What is the largest area the rectangle can have?

### Step 1

\(A = 2xy\)

\(y = 3 - x^2\)

Substitute \(y\) into the area equation: \(A = 2x(3-x^2)=6x-2x^3\)

### Step 2

Maximize the area by taking the first derivative and set it equal to \(0\).

\(A’ = 6 - 6x^2 = 0 \rightarrow x = 1\)

\begin{array}{|c|c|c|} \hline +& 0 & - \\ \hline & x = 1 \end{array}

\(A’(0) = 6 - 6(0)^2 = 6 \rightarrow \text{positive value}\)

\(A’(2) = 6 - 6(2)^2 = -18 \rightarrow \text{negative value}\)

### Step 3

Substitute \(x = 1\) into \(y\)

\(y = 3 - 1^2 = 2\)

\(\text{Maximum Area} = 2xy = 2(1)(2) = 4\)

## Example 2

Find the dimensions of a right circular cylindrical can (with bottom and top closed) that has a volume of 1 liter and that minimizes the amount of material used. (*Note*: One liter corresponds to \(1000 \ \mathrm{cm}^3\).

### Step 1

\(V = \pi r^2h = 1000 \ \mathrm{cm}^3 \rightarrow h = \frac {1000} {\pi r^2} \)

\(\text{Surface Area} = S_\text{top} + S_\text{bottom} + S_\text{side}\)

\(\text{Surface Area} = \pi r^2 + \pi r^2 + 2\pi r h\)

### Step 2

Minimize the surface area

\(\text{Surface Area} = \pi r^2 + \pi r^2 + 2\pi r h\)

\(= \pi r^2 + \pi r^2 + 2\pi r (\frac {1000} {\pi r^2})\)

\(= 2 \pi r^2 + \frac {2000} {r} \)

\(S’ = 4\pi r - \frac {2000} {r^2}\)

\(= \frac {4\pi r^3 - 2000} {r^2} = 0 \)

\(4\pi r^3 - 2000 = 0 \rightarrow r^3 = \frac {2000} {4\pi}\)

\(r = 5.42\)

\begin{array}{|c|c|c|} \hline -& 0 & + \\ \hline & r = 5.42 \end{array}

\(S’(-1) \rightarrow \text{negative value}\)

\(S’(6) \rightarrow \text{positive value}\)

### Step 3

Substitute back in \(r = 5.42\) to Volume equation to get the value of \(h\)

\(h = \frac {1000} {\pi (5.42)^2} = 10.84 \ \mathrm{cm} \)