Linear demand and cost—Quadratic revenue and profit:
A manufacturer has weekly fixed costs totaling $160. Each unit costs an additional $12 to produce.
Find the cost function: C ( x )  160  12 x
The demand quantity, x , is 0 if the selling price is at or above $30. Each decrease of $1 in the price
increases demand quantity by 5 units.
Find the demand price function: slope  
1
  0 . 2 , y-intercept = 30 , p   0 . 2 x  30
5
What is the maximum demand quantity even if the product is free?
This is the x-value when p=0. 0   0 . 2 x  30 ,
x
30
 150
.2
Find the revenue function: R ( x )  xp ( x )  x (  . 2 x  30 )   . 2 x  30 x
2
Find the maximum revenue: method 1 The zeros of revenue at x=0 and x=150. So the revenue is a max
when x=75. R ( 75 )  75 (  . 2 ( 75 )  30 )  1025
method 2 h  
b

2a
30
 .4
R ( 75 )  1025
 75
P ( x )  R ( x )  C ( x )   . 2 x  30 x  (160  12 x )
2
Find the profit function:
P ( x )   . 2 x  18 x  160
2
Find the break even quantities: Use the calculator or factor the profit function or put it into vertex form
and solve. The zeros are 10 and 80.