An insurance company's monthly revenue from x policies is R(x)=125x and monthly cost is C(x)=100x+5000. (1) The break-even point is found by setting R(x)=C(x) and solving for x, which is x=200 policies. (2) The revenue and cost functions are graphed on the same axes. (3) At x=100,000 policies, the estimated revenue is $12,500 and cost is $15,000.

1. SEC. 3.3 P. 143 #33
HOW TO SOLVE IT

2. THE ACTUAL PROBLEM
• An insurance company claims that for x thousand policies, its monthly
revenue in dollars is given by R(x)=125x and its monthly cost in dollars is given
by C(x)=100x+5000.
• (a) Find the break-even point.
• (b) Graph the revenue and cost equations on the same axes.
• (c) From the graph, estimate the revenue and cost when x=100 (100,000
policies).

3. (A) FIND THE BREAK-EVEN POINT.
• In order to find a break-even point you must set your R(x) equal to C(x),
which should look like
125x=100x+5000

4. (A) FIND THE BREAK-EVEN POINT
• Next you want to move all your “x’s” to one side
• Original equation: 125x=100x+5000
• Equation after moving all the “x’s” to one side (subtract 100)
•25x=5000

5. (A) FIND THE BREAK-EVEN POINT
• Now you want to solve for X (divide 25 by both sides)
•25x=5000
• So now your answer will look like
• X=200

6. (A) FIND THE BREAK-EVEN POINT
• Your breakeven point is at 200
• So for 200 insurance policies being sold, the break even
point will be met

7. (B)GRAPH THE REVENUE AND COST EQUATIONS ON THE SAME AXES

8. (C) FROM THE GRAPH, ESTIMATE THE
REVENUE AND COST WHEN X =100
(100,000 POLICIES)
• When X is 100 for revenue it equals 12500
• When X is 100 for Cost it equals 15000

9. DISCUSSION
• Overall this problems wasn’t really hard, I just didn’t find the right kind of
graph to make it easier on myself. All of the work was achievable.