Question

3. The marketing research department for a company that manufacturers and sells notebook computers established the following revenue and cost functions:
R(x) = x(2000 – 60x)
C(x) = 4000 + 500x,
where x is thousands of computers, and R(x) and C(x) are in thousands of dollars. Both
functions have the domain 1 ≤ x ≤ 25.
(1) Form a profit function P, and graph R, C, and P in the same rectangular coordinate system.
(2) Discuss the relationship between the intersection points of the graphs of R and C and the x intercepts of P.
(3) Find the x intercepts of P to the nearest hundred computers. Find the break-even points.
(4) Refer to the graph drawn in part (1). Does the maximum profit appear to occur at the same output level as the maximum revenue? Are the maximum profit and maximum revenue equal? Explain.  

Answer 1

R(x) = x(2000 – 60x)
C(x) = 4000 + 500x
P(x) = R(x) -C( x) = x(2000 – 60x) - (4000 + 500x) = -60x² + 1500x - 4000

(1)

(2)

As seen in the graph, the intersection point of R(x) and C(x) is the point where Profit (P(x)) becomes zero, (or the x-intercept are the points where the graph of R(x) and C(x) intersect).

(3) The x-intercepts are itself the breakeven points.
x-intercepts are 3.035 and 21.965.

(4) No. Mathematically, since these two are different equations, their maximum points might not be the same which is the case here. Also, since the cost involves both fixed cost and variable cost, hence the maximum point of P(x) and R(x) are different.

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