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Worksheet 10 Solutions Applications of Quadratic Functions 1) The supply function for a product is given by p = q2 + 300 and the demand function is given by p + q = 410. Find the equilibrium quantity and price. Rewrite p + q = 410 as p = 410 - q Equilibrium is when supply = demand q2 + 300 = 410 - q Subtract 410 from both sides q2 – 110 = - q Add q to both sides q2 + q – 110 = 0 Factor (q + 11) (q – 10) = 0 So q = -11, 10 but can’t have a negative so q = 10 Plug in q = 10 into p = q2 + 300 p = (10)2 + 300 = 400 Equilibrium point is (10, $400). 2) If the total costs for a product are given by C(x) = 1760 + 8x + 0.6x2 and total revenue is given by R(x) = 100x – 0.4x2, find the break-even point. Break even is when R=C 100x – 0.4x2 = 1760 + 8x + 0.6x2 Subtract 1760, 8x and .6x2 from both sides -x2 + 92x – 1760 = 0 Use the quadratic formula a = -1 b = 92 c = -1760 x b b 2 4ac 92 922 4(1)(1760) 2a 2( 1) 92 1424 2 92 1424 27.132 28, 2 So 92 1424 x 64.868 65 2 x 3) The profit function for a firm is P(x) = 80x – 0.1x2 – 7000. Find the number of units at which the maximum profit is achieved, and find the maximum profit. Find the vertex a = -.1 b = 80 x b 80 400 2a 2(.1) Plug in x=400 80(400) – 0.1(400)2 – 7000 = 9000 Must sell 400 units to earn a maximum profit of $9,000. 4) A certain company has fixed costs of $15,000 for its product and variable costs given by 140 + 0.04x dollars per unit, where x is the total number of units. The demand for the product is given by p = 300 – 0.06x. a. Find the cost and revenue equation. C(x) = 15000 + (140 +0 .04x) = 15000 +140x + 0.04x2 R(x) = px = (300 – 0.06x) x = 300x – 0.06x2 b. When will the company break-even? Break even is when R=C 300x – 0.06x2 = 15000 + 140x + 0.04x2 Subtract 300x and Add 0.06x2 from both sides 0 = 15000 -160x + x2 0 = (x – 150) (x – 100) So x=1500, 100 c. Find the price that will maximize revenue. R(x) = 300x – 0.06x2 Find the vertex a = -.06 b = 300 x b 300 2500 2a 2(.06) Plug in x=2500 to p = 300 – 0.06x. 300 – 0.06(2500) = 150 A price of $150 will maximize revenue. d. Find the profit equation. P(x) = R – C = (300x – 0.06x2) – (15000 +140x + 0.04x2) = -.1x2 +160x – 15000 e. What is the production level that will maximize profit and what is the maximum profit? Use P(x) = -.1x2 +160x – 15000 Find the vertex a = -.1 b = 160 x b 160 800 2a 2(.1) plug in x=800 -.1(800)2 +160(800) – 15000 = 49000 Must sell 800 units to earn a maximum profit of $49,000.