Chapter 6.4, Problem 41PPS

To determine

To graph: for the given function and most profitable price

Answer to Problem 41PPS

The most profitable price is $\$\text{1}.\text{25}$

Explanation of Solution

Given information:

The weekly earning can be represented as $ƒ\left(\text{x}\right)$

  $ƒ\left(\text{x}\right)=$ $-5x^{2}100x+2625$

Where ,x is the number of $\$0.0\text{5}$ increases

Calculation:

The values can be found by substituting the values in the given function

Therefore using the tables to the function

$x$	$ƒ\left(\text{x}\right)$
$0$	$2625$
$1$	$2720$
$2$	$2805$
$10$	$3125$
$20$	$2625$

To graph:

Using MAPLE to graph the given expression

  

The initial rate of beverage is given as $\$0.\text{7}$ per can and increase in rate will be $\$0.0\text{5}$ per can

Therefore considering the graph ,in this graph profit will be maximum at the point $\text{x}=\text{1}0$

So the total increment at $\text{x}=\text{1}0$ will be calculated as

Total increment $=x\times 0.0\text{5}$

  $\begin{array}{c}\text{Increament}=\text{1}0\times 0.0\text{5}\\ =50.5\end{array}$

The initial price is

  $IP=\$0.\text{75}$

Most profitable price is

  $\$0.\text{5}+\$0.\text{75}=\$\text{1}.\text{25}$

Therefore, the most profitable price is $\$\text{1}.\text{25}$