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.
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Answer:
The fraction of total interest owed is
The value of it to the nearest tenth is 73.1%
Step-by-step explanation:
Let us solve it using the rule of 78
∵ The loan = 12 month
∵ The numerator is after 6 month
- Substitute n by 1 in each bracket
∴ The numerator = {(1 + 11) + (1 + 10) + (1 + 9) + (1 + 8) + (1 + 7) + (1 + 6)}
∴ The numerator = {12 + 11 + 10 + 9 + 8 + 7}
∴ The numerator = {57}
∴ The numerator of the fraction is 57
∵ The denominator = {(n) + (n + 1) + (n + 2) + .......... + (n + 11)}
- substitute n by 1 in each bracket
∴ The denominator = {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12}
∴ The denominator = {78}
∴ The denominator of the fraction is 78
The fraction of total interest owed is
Now find the value of the fraction
∵ The fraction =
∴ Its value = 0.7307692308 × 100%
∴ Its value = 73.07692308%
- Round it to the nearest tenth
∴ Its value = 73.1%
The value of it to the nearest tenth is 73.1%