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.

Short 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-by-step solution

Given information

The equation can be given as

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

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

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