Q.92 - Precalculus Enhanced with Graphing Utilities

Question

The daily revenue R achieved by selling x boxes of candy is figured to be $R\left(x\right)=9.5x-0.04x^2$ The daily cost C of selling x boxes of candy is $C\left(x\right)=1.25x+250$.

(a) How many boxes of candy must the firm sell to maximize revenue? What is the maximum revenue?

(b) Profit is given as P(x) = R(x) - C(x). What is the profit function?

(c) How many boxes of candy must the firm sell to maximize profit? What is the maximum profit?

(d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a quadratic function is a reasonable model for revenue.

Step-by-Step Solution

Answer

a)To maximize the revenue 119 boxes should be sold and revenue is $564

b)The profit equation is $P\left(x\right)=-0.04x^2+8.25x-250$

c)to maximize the profit 103 boxes should be sold and the profit is $175

d)As the function of profit depends on the cost function and it does not depend on revenue function Hence the maximum values are different  

Step 1: Given information

The equation can be given as

$$\begin{gathered} R\left(x\right) = 9.5x - 0.04x^2 \\ C\left(x\right) = 1.25x + 250. \end{gathered}$$

Step 2: Find the vertex

in the revenue equation the coefficient of $x^2$ term is negative. hence the parabola opens downwards. the maximum value is at the vertex

The vertex can be given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=\frac{9.5}{0.08} \\ x=118.75 \end{gathered}$$

The corresponding revenue is 

$$\begin{gathered} R(118.75)=9.5(118.75)--0.04{(118.75)}^2 \\ R(118.75)=564.06 \end{gathered}$$

Part b) Step 1: Find the profit function by substituting the corresponding function in the equation

We get

$$\begin{gathered} P\left(x\right)=9.5x-0.04x^2-1.25x-250 \\ P\left(x\right)=-0.04x^2+9.25x-250 \end{gathered}$$

Part c) Step 1) Find the vertex of the quadratic equation of profit

As the coefficient of $x^2$term is negative the parabola opens downward hence the maximum value is at the vertex

The vertex can be given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=\frac{8.25}{0.08} \\ x=103.125 \end{gathered}$$

The maximum profit is 

$P(103.125)=175.39$

Part d) Step 1: Explanation

As the function of profit depends on the cost function and it does not depend on revenue function Hence the maximum values are different  

Step 6: Conclusion

a)To maximize the revenue 119 boxes should be sold and revenue is $564

b)The profit equation is $P\left(x\right)=-0.04x^2+8.25x-250$

c)to maximize the profit 103 boxes should be sold and the profit is $175

d)As the function of profit depends on the cost function and it does not depend on revenue function Hence the maximum values are different