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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.
Answer:
x = ± 5
Step-by-step explanation:
Given
x² - 4x - 25 = - 4x ( add 4x to both sides )
x² - 25 = 0 ( add 25 to both sides )
x² = 25 ( take the square root of both sides )
x = ± = ± 5
Area of the square = 72.25
Side = 8.5 in
Perimeter = 4 × side
Perimeter = 4 × 8.5 in
Perimeter = 34 in