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A manufacturer finds that it costs him x^2 +5x+7 dollars to produce x tons of an item. At production levels above 3 tons, he must hire additional workers, and his costs increased by 3(x-3) dollars on his total production. If the price he recieves is $13 per ton regardless of how muvch he manufactures and he has a plant capacity of ten tons what level of output maximizes his profits?

The number x tons produced needs to be at least 0 but no more than 10 (you can't store more than 10 tons) AND you have the additional stipulation that when x>3 production costs go up you have a split domain of 0<=x<=3 and 3<x<=10

let x = # of tons of items produced

profit is equal to revenue minus costs

P(x) = R(x) - C(x) ---> P(x) = 13x -(x^2+5x+7) 0<=x<=3

= 13x -(x^2+5x+7+3(x-3)) 3<x<=10

For 0<=x<=3,

P(x) = 13x -(x^2+5x+7)--> -x^2 + 8x -7 and P'(x) = -2x+8

-2x+8 = 0 --> x = 4 and 4 is not in 0<=x<=3

For 3<x<=10

P(x) = 13x -(x^2+5x+7+3(x-3))--> -x^2+5x+2 and P'(x) = -2x + 5

-2x+5 = 0 --> x = 5/2 and 5/2 is not in 3<x<=10

There are techniques without calculus that say that a max or a min must occur where two of the lines when graphed intersect. Your lines are x = 0, x = 3, and x = 10 as well as the parabolas from each part: -x^2 + 8x -7 and -x^2+5x+2

You might try graphing these and see what point gives the largest y value

The two parabolas intersect at x = 3, where one is increasing and the other is decreasing so at x = 3 you have 3 tons as a max.

Professor Wallin

Advanced Math

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