College Algebra 9th txtbk

SECTION 3–4 Quadratic Functions 213
Step 4. We need to identify intervals where f(x) 0. From the graph we see that
f(x) 0 for x 2 312 and for x 2 312. Returning to the original
inequality, the solution to
x 2 4x 14
is
(, 2 312] [2 312, )
MATCHED PROBLEM 8 Solve: x 2 6x 6
EXAMPLE 9 Break-Even, Profit, and Loss
Table 2 Price–Demand Data
Weekly Sales Price
(in gallons) per Gallon
1,400 $43.00
2,550 $37.25
3,475 $32.60
4,856 $25.72
5,625 $21.88
6,900 $15.50
Table 2 contains price–demand data for a paint manufacturer. A linear regression model for
this data is
p 50 0.005x
Price–demand equation
where x is the weekly sales (in gallons) and $p is the price per gallon. The manufacturer
has weekly fixed costs of $58,500 and variable costs of $3.50 per gallon produced.
(A) Find the weekly revenue function R and weekly cost function C as functions of the
sales x. What is the domain of each function?
(B) Graph R and C on the same coordinate axes and find the level of sales for which the
company will break even.
(C) Describe verbally and graphically the sales levels that result in a profit and those that
result in a loss.
SOLUTIONS
(A) If x gallons of paint are sold weekly at a price of $p per gallon, then the weekly
revenue is
R xp x(50 0.005x) 50x 0.005x 2
Since the sales x and the price p cannot be negative, x must satisfy
x 0 and p 50 0.005x 0
0.005x 50
x
The revenue function and its domain are
50
0.005 10,000
Subtract 50 from both sides.
Divide both sides by 0.005
and reverse the inequality.
Simplify.
R(x) 50x 0.005x 2
0 x 10,000
The cost of producing x gallons of paint weekly is
C(x) 58,500 3.5x x 0
(B) The graph of C is a line and the graph of R is a parabola opening downward. Using
the vertex formula,
The vertex is (5,000, 125,000).
Fixed costs $3.50 times
number of gallons
x b
2a 50
2(0.005) 5,000
R(5,000) 50(5,000) 0.005(5,000) 2 125,000