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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.