​(a) How many wristwatches 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?

Question

The monthly revenue R achieved by selling x wristwatches is figured to be

Upper R left parenthesis x right parenthesis equals 79 x minus 0.2 x squared .R(x)=79x−0.2x2.  

The monthly cost C of selling x wristwatches is Upper C left parenthesis x right parenthesis equals 32 x plus 1650.C(x)=32x+1650.

​(a) How many wristwatches must the firm sell to maximize​ revenue? What is the maximum​ revenue?

​(b) Profit is given as

Upper P left parenthesis x right parenthesis equals Upper R left parenthesis x right parenthesis minus Upper C left parenthesis x right parenthesis .P(x)=R(x)−C(x).

What is the profit​ function?

​(c) How many wristwatches 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.

Answer 1

(a) Rewrite:

R(x)=0.2((79/0.4)²-((79/0.4)²-79x/0.2+x²))=0.2((79/0.4)²-(x-79/0.4)²).

This rearrangement shows that when x=79/0.4, R(x) has a maximum value of 0.2(79/0.4)²=7801.25.

However, x=79/0.4=197.5 wrist watches. R(197)=R(198)=7801.20. The sale of 197 or 198 watches raises 7801.20 in revenue.

(b) P(x)=R(x)-C(x)=79x-0.2x²-32x-1650=47x-0.2x²-1650 is the profit function.

(c) To maximise profit x=47/0.4=117.5, so P(117)=P(118)=1111.20. 117 or 118 watches need to be sold.

(d) (a) doesn’t take into account the cost of making the watches. A quadratic function provides a target for the vender and allows the firm to plan production more efficiently. Also, it is easier to work out the target when a quadratic function applies. A linear equation cannot fix a target, because profit just increases with sales. More complex functions (polynomials with higher degree than 2) are more difficult to solve to find targets, and may generate more than one target.