Applications of Differentiation - Optimization

guest:

Dear All,

Please help me about the Applications of Differentiation - Optimization below :

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The demand function is : P = 400 - 2Q

Average Cost : AC = 0.2Q + 4 + (400/Q)

Questions :

a. Determine how many units of a product that must be produced if the company wants to get the maximum profit?
b. Determine the selling price when the profit is maximum.
c. Determine the maximum profit that can be earned by the company.



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Answer :

P = 400 – 2Q

TR = P.Q = (400 – 2Q)Q = 400Q – 2Q²

TC = AC.Q = (0.2Q + 4 + (400/Q))Q = 0.2Q² + 4Q + 400

Profit = TR – TC
= (400Q – 2Q²) – (0.2Q² + 4Q + 400)
= – 2.2Q² + 396Q – 400

I don't know the next step, please help me..

Thank you very much for your answer..

Ok, so far, so good.
a. Determine how many units of a product that must be produced if the company wants to get the maximum profit?
The profit function is a concave down parabola (parabola because the biggest power of Q is 2, and concave down because the coefficient of Q ² is negative (-2.2))
The maximum point wil be when the gradient of the tangent is 0
That is when dProfit/dQ=0 [Note: Marginal profit = MProfit = dProfit/dQ]
So Mprofit = -4.4Q + 396 (I have just differentiated profit)
Solve Mprofit = 0 and you will get the quantity that will give the maximum profit.

b. Determine the selling price when the profit is maximum.
Substitute the Q that you just got into the P function to get the selling price per unit

c. Determine the maximum profit that can be earned by the company.
substitute the Q that you just got into the Profit function.

Dear Melody,

Is this correct ?

a. -4.4Q + 396 = 0
-4.4Q = -396
Q = 90

b. P = 400 - 2Q
= 400 - (2 x 90)
= 400 - 180
= 220

c. Profit = - 2.2Q2 + 396Q - 400
= - (2.2 x 90 ²) + (396 x 90) - 400
= - 17820 + 35640 - 400
= 17420

Thank you very much..

Yes, I believe that it is.

Dear Melody,

Thank you very much..