Answer to Question #281748 in Calculus for Rucha

Question

The marketing department of Spager Ltd estimated that if the selling price of product is set at $15 per unit then the sales will be 50 units per week, while, if the selling price is set at $20 per unit, the sales will be 30 units per week. Assume that the graph of this function is linear. The production department estimates that the variable cost will be $5 per unit and that the fixed cost will be $50 per week, and special cost are estimated as $0.125x2, where x is the quantity of output. All production is sold.

(a) Show that the relationship between price (Pr) and quantity sold (x) , are given by the equation Pr = 27.5 - 0.25x.

(b) Find the revenue function, R.

(c) Find the total cost function (C).

(d) Advise the company on production and pricing policy if it wishes to maximize profits, and find the maximum profit.

Answer 1

(a)Let's assume the graph of the function is linear;

$Pr(x)=mx+b$

$Pr(50)=m(50)+b=15$

$Pr(30)=m(30)+b=20$

$50m-30m=15-20\implies m=-0.25$

$-0.25(30)+b=20\implies b=27.5$

Thus;

Pr=27.5-0.25x

(b)R(x)=xPr(x)

R(x)=x(27.5-0.25x)

R(x)=27.5x-0.25x²

(c)C(x)=5x+50+0.125x²

(d)P(x)=R(x)-C(x),$x\ge 0$

=27.5x-0.25x²-(5x+50+0.125x²)

=22.5x-0.375x²-50

P'(x)=22.5-0.75x

For critical numbers P'(x)=0

22.5-0.75x=0

x=30

If 0<x<30, P'(x)>0 P(x) increases

If x>30, P'(x)<0 P(x) decreases

The function P'(x) has a local maximum at x=30. Since the function P(x) has the only extreme the function P(x) has the absolute maximum for $x\ge 0$ at x=30

P(30)=22.5(30)-0.375(30²)-50=287.50

Hence maximum profit will be $287.50 at x=30 units per week.