Answer:
$545
Explanation:
Finding the price that maximizes profit requires several steps. We need the demand function that tells quantity sold in terms of selling price. With this, we can express the cost and revenue functions in terms of selling price. The profit will be the difference between revenue and cost.
__
demand
The demand (q) is said to be a linear function of price (p). The equation for that can be written ...
q(p) = (q2 -q1)/(p2 -p1)(p -p1) +q1 . . . . . for two points (p1, q1) and (p2, q2)
Using the given demand points, this can be ...
q(p) = (1900 -2600)/(575 -470)(p -470) +2600
q(p) = -20/3(p -470) +2600
q(p) = -20/3p +5733 1/3
__
cost
The cost function in terms of selling price can be found by substituting the q(p) equation for x in c(x). For the purpose of finding the maximum profit, we only need the marginal cost function:
c(q(p))' = c'(q(p))·q'(p) = (608 -0.18(q(p)))·(-20/3) = (608 -0.18(-20/3p +5733 1/3))(-20/3)
c'(p) = -8p +2826 2/3
__
revenue
As with cost, the function we need for finding the maximum profit is the marginal revenue function. Revenue is the product of price and demand, so the marginal revenue is ...
r(p) = p·q(p)
r'(p) = q(p) +p·q'(p) = (-20/3p +5733 1/3) +p(-20/3)
r'(p) = -40/3p +5733 1/3
__
maximum profit
The maximum profit is found where the marginal profit is zero. In turn, that is found where the difference between marginal revenue and marginal cost is zero:
P = R -C
P'(p) = 0 = r'(p) -c'(p) = (-40/3p +5733 1/3) -(-8p +2826 2/3)
0 = -16/3p +2906 2/3
16p = 8720 . . . . . multiply by 3, add 16p
p = 545 . . . . . . . price for maximum profit
The price per set that maximizes profit is $545.00.