Question

3. A certain company has fixed costs of $15,000 for its product and variable costs is given by 140 +0.04x dollars per unit where x is the total number of units. The selling price p of the product is given by 300 -0.06x dollars per unit. (6 pts) a) Set up the functions for total cost and total revenue. b) Find the minimum break even quantity. c) Set up the profit function and compute the number of units that will maximize profit. d) What selling price will maximize the profit? Then compute the maximum profit.

Answer 1

giny Fixed cost = $15,000 Variable Cost = 140+ 0.04 a doller/unit Selling Price = 3 00-0,06x dolles/unit let W unit produce & sell total cost = fined cost t Variable cost Con= 15000 + (14o tooke) .se ((x) = 15000 + 140x to.o¢ x² stotal Revenue a Belling Price) x quantity = (300x0.06x). 2 I Rex) = 3008-0.868² l dolley at minimum break event Capacity I total cost a total sells 15000+140 x 70.04 x2 = 3008 010682

0.1x² - 1 box + 15ooooo na 160 +16624(0-1)(15000) - 210.4) * = 160+ 140 3002 0 2 u=15oo, 100 So, Minimum Break even capacity (se=100 units) @ Pərv it fоr ) = fеwhua - Cos+ = (300X-0.06%?) – ( 15000 + 140x + oogat" Pexeso -0.4x2 +10% - 15000 dpa (-0.12€+ 1602 m (5000) = -0.2x+leo To maximize profit of a no 0.26 +16020 => 2 = 800 da da'

and derivative test as of 1 (0.22 +160) = -0.250 -0,2 a for maximum profit X=800 units So, Maxima at 82 800 profit maximum at x=800 So, selling price = 300 -0.064 doller/ unit at x=800 P = 300 - 0.06(800) Selling P = 252 doller lunit 1 pilce Profit (P) = -0.1x? +1684- 19000 at x=800 Pmax. = P(800) = -0.1(800)+1601860)41 so0o Pmax= 49,000 dollez O