1 - Radford



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

[pic]

So [pic]

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

[pic]

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

[pic]

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

[pic]

plug in x=800

-.1(800)2 +160(800) – 15000 = 49000

Must sell 800 units to earn a maximum profit of $49,000.

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