(1 pt) A company sells wrought iron umbrella trees. This company has fixed costs of $100,000/month. The cost in labor and material is $15 per umbrella sold per month. To sell x items, the price must be set at $50-0.3x, where x is in thousands of umbrellas sold per month. What price will maximize profit?
price=
Calculus problem?
$22.38
Reply:Total cost per month if x umberellas are sold per month is
C(x) = 100,000 + 15x
The Revenue generated by selling x umberalls per month is
product of umberalls and price of each umberalla
so R(x) = x(50 - 0.3x)
R(x) = 50x - 0.3x^2
profit = R(x) - C(x)
so P(x) = 50x - 0.3x^2 - (100,000 + 15x)
P(x) = -0.3x^2 + 35x -100000
differentiate with respect to x
dP/dx = -0.6x + 35
to get maximum profit
dP/dx should be zero
so -0.6x + 35 = 0
0.6x = 35
x = 35/0.6 = 58.33
since price of umberalla, p is 50 - 0.3x
p = 50 - 0.3(58.33) = $ 32.5
At the price of $ 32.5 per umberalla , the profit will be maximum
Reply:Profit = income - costs
Income = number of umbrellas * price
= 1000x(50-0.3x)
Costs = 100,000 + 15000x
So profit = 1000x(50-0.3x) -100000 - 15000x
= 35000x - 300x^2 - 100000
Want to maximise this, so differentiate it,
d(profit)/dx = 35000 - 600x
Solve d(profit)/dx = 0 =%26gt; x = 58.333333333
So price = 50 - 0.3*x = $32.50
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