SOLUTION: I'm having a hard time setting up this equation... A manufacturer finds that for the first 300 units of its product that are produced and sold, the profit is $60 per unit. The pro

Question 161544: I'm having a hard time setting up this equation...
A manufacturer finds that for the first 300 units of its product that are produced and sold, the profit is $60 per unit. The profit on each of the units beyond 300 is decreased by $0.10 times the number of additional units sold. What level of output will maximize profit?
I can solve these when I am givin the function but I need help setting this one up. Then I will be more well equipped to solve these in the future.
Thanks to all the tutors out there, your help is greatly appreciated.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A manufacturer finds that for the first 300 units of its product that are produced and sold, the profit is $60 per unit. The profit on each of the units beyond 300 is decreased by $0.10 times the number of additional units sold. What level of output will maximize profit?
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For the 1st 300, Profit = 300*60
For all over 300, the profit is $60 - $0.10 times the number sold, = $60 - $0.10*(x-300).
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Overall then,
Profit = 300*60 + (x-300)*(60 - 0.1(x-300))
P = 18000 + (x-300)*(60 - 0.1x + 30)
P = 18000 + (x-300)*(-0.1x + 90)
P = 18000 -0.1x^2 + 120x - 27000
P = -0.1x^2 + 120x - 9000
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To find the max, set the 1st derivative = 0
0 = -0.2x + 120
0.2x = 120
x = 600

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