The quantity demanded x for a product is inversely proportional to the cube of the price p for p > 1. When the price is $10 p
er unit, the quantity demanded is 64 units. The initial cost is $150 and the cost per unit is $2. What price will yield a maximum profit?
1 answer:
Q(p) = k/p^3 . . . . . . . . we want to find k
q(10) = k/10^3 = 64
k = 64,000
Revenue = q(p)*p = 64000/p^2
Cost = 150 +2q = 150 +2*64000/p^3
Profit = Revenue -Cost = 64000/p^2(1 -2/p) -150
Differentiating to find the maximum profit, we have
.. dProfit/dp = -2(64000/p^3) +6(64000/p^4) = 0
.. -1 +3/p = 0
.. p = 3
A price of $3 per unit will yield a maximum profit.
q(10) = k/10^3 = 64
k = 64,000
Revenue = q(p)*p = 64000/p^2
Cost = 150 +2q = 150 +2*64000/p^3
Profit = Revenue -Cost = 64000/p^2(1 -2/p) -150
Differentiating to find the maximum profit, we have
.. dProfit/dp = -2(64000/p^3) +6(64000/p^4) = 0
.. -1 +3/p = 0
.. p = 3
A price of $3 per unit will yield a maximum profit.
You might be interested in
Answer:
The domain is all real numbers. The range is {y|y ≤ 16}
Step-by-step explanation:
we have
This is a the equation of a vertical parabola open downward
The vertex is a maximum
The vertex is the point (-1,16)
see the attached figure
therefore
The domain of the function is all real numbers ----> interval (-∞,∞)
Te range of the function is
All real numbers less than or equal to 16 ----> interval (-∞,16]
