Chapter 12.3, Problem 72E

To determine

To show: that the profit is increasing on the interval $\left[0,20\right]$ and decreasing on the interval $\left[40,60\right]$

Explanation of Solution

Given information:

The profit $P$ that a company makes depends on the amount $x$ the company spends on advertisting. The profit function is

  $P\left(x\right)=200+30x-0.5x^2$

Using your knowledge of the slopes of tangent lines.

Calculation:

Since, the slope of the graph of $f$ at point $\left(x,f\left(x\right)\right)$ is equal to the slope of its tangent line at $\left(x,f\left(x\right)\right)$ is:

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Find slope at point $\left(0,20\right)$

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{200+30\left(0+h\right)-0.5{\left(0+h\right)}^2-200-30\times 0+0.5\times 0^2}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(30-0.5h\right)\cancel{h}}{\cancel{h}}\\ =30\end{array}$

Since, the slope is positive, So, the height of the ball is increasing on the interval $\left(0,20\right)$ .

Find slope at point $\left(40,60\right)$ :

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(40+h\right)-f\left(40\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{200+30\left(40+h\right)-0.5{\left(40+h\right)}^2-200-30\times 40+0.5\times {40}^2}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(30-0.5h-40\right)\cancel{h}}{\cancel{h}}\\ =-10\end{array}$

Since, the slope is negative. So the height of the ball is decreasing on the interval $\left(40,60\right)$ .