word problems

Question

A firm has demand schedule p = 150 - q and cost function C (q) = 2000 + 40q a) Give the firm's revenue function R (q). b) What level of production will maximize the firm's revenues? c) Find the profit function P (q). d) Find the firm's break-even points. e) What level of production will maximize the firm's profits

Answer 1

p=150-q, c=2000-40q

Revenue = R(q) = pq = (150-q)q = 150q -q^2

R' = 150-2q = 0

q = 150/2 = 75 = revenue maximizing quantity

p = 150-75 = revenue maximizing price

Profit =P= R(q)-C(q) = 150q -q^2 -2000 +40q = -q^2 +190q -2000

P'(q) = -2q + 190 = 0

q = 190/2 = 95 = profit maximizing sales

p = 150-95 = $55= profit maximizing price

max Profit = 95x55 = $5,225

break even points are when profit = 0 = -q^2 +190q -2000

q^2 -190q +2000 = 0

complete the square or use the quadratic formula

q= 95+/-5sqr281= about 11 or 179

p= 95 +/-5sqr281= about $11 or $179