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.

(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.

answered Nov 19, 2017 by Top Rated User (582,840 points)