12th Lecture

Singular and Stationary Points

To find stationary points, set the derivative $f'(x) = 0$. Singular points occur where $f'(x)$ is undefined, but $f(x)$ is defined.

Example Function

Given $f(x) = x + \frac{2}{x}$, its derivative is $f'(x) = 1 - \frac{2}{x^2} = \frac{x^2 - 2}{x^2}$.

• Stationary points: $x = \pm \sqrt{2}$.

• The function $f(0)$ is undefined, so $x=0$ is not a singular point for $f(x)$.

• Relative maximum at $x = -\sqrt{2}$, relative minimum at $x = \sqrt{2}$.

Using Derivatives and Graphing Tools

Graph $f(x)$ and $f'(x)$ to locate extreme points. Stationary points occur where $f'(x) = 0$. Singular points occur where $f'(x)$ has a vertical asymptote but $f(x)$ is defined.

Average Cost Minimization

Average cost $A(x) = \frac{C(x)}{x}$, where $C(x)$ is the total cost. To minimize average cost, find where $A'(x) = 0$.

Example: Hockey Jerseys

Cost function: $C(x) = 2000 + 10x + 0.2x^2$. Average cost: $A(x) = \frac{2000}{x} + 10 + 0.2x$. Derivative: $A'(x) = -\frac{2000}{x^2} + 0.2$.

• Set $A'(x) = 0$ to find stationary points.

• In this case, $x = 100$ minimizes the average cost, with an average cost of $50 per jersey.

Optimization Problem: Maximizing Area

Objective: Maximize area $A = xy$. Constraint: Limited fencing, $x + 2y = 100$.

• Express $x$ in terms of $y$: $x = 100 - 2y$.

• Substitute into the area equation: $A(y) = (100 - 2y)y = 100y - 2y^2$.

• Find the derivative: $A'(y) = 100 - 4y$.

• Set $A'(y) = 0$ to find stationary points: $y = 25$.

• Then, $x = 100 - 2(25) = 50$.

Dimensions for maximum area: 50 feet across and 25 feet deep.