Chapter 4.4, Problem 61E

To determine

To find: To find the profit is maximum at the point.

Answer to Problem 61E

The profit is maximum at the point is $(0,0),(3,0)$ .

Explanation of Solution

Given:

The cost is $c(x)=x^3-6x^2+15x$ and the revenue is given by $r(x)=6x$ .

Formula used: $p(x)=r(x)-c(x)$

Calculation:

To find the maximum profit,

  $\begin{array}{l}p\text{'}(x)=0\\ p\text{'}(x)=r\text{'}(x)-c\text{'}(x)=0\\ r\text{'}(x)=c\text{'}(x)\end{array}$

Profit will be maximum if,

  $\begin{array}{l}p\text{'}\text{'}(x)<0\\ (r-c)\text{'}\text{'}(x)<0\end{array}$

  $\begin{array}{c}r(x)=6x\\ r\text{'}(x)=6\\ c(x)=x-6x^2+15x\\ c\text{'}(x)=3x^2-12x+15\\ c\text{'}(x)=3(x^2-4x+5)\\ c\text{'}(x)=3(x-5)(x+1)\end{array}$

  $\begin{array}{c}2=x^2-4x+5\\ x^2-4x+3=0\\ (x-3)(x-1)=0\\ x=1,3\end{array}$

Here $p\text{'}(x)<0$ for $0<x<1$ and $p\text{'}(x)>0$ for $1<x<3$ also $p\text{'}(x)<0$ for $x>3$

Absolute maxima occurs at endpoints $x=0$ and critical point $x=3$ .

  $p(0)=p(3)=0$ The function has absolute maximum at the endpoints $(0,0),(3,0)$ .

Graphically supporting