If c ( x ) = 11000 + 500 x − 3.6 x 2 + 0.004 x 3 is the cost function and p ( x ) = 1700 − 9 x is the demand function, find the production level that will maximize profit. (hint: if the profit is maximized, then the marginal revenue equals the marginal cost.)

Step-by-step explanation:

The cost function = C(x)

The demand function = P(x)

C(x) = 11000 + 500x - 3.6x^2 + + 0.004x^3

P(x) = 1700 - 9x

Differentiate C(x) with respect to x

C'(x) = 500 - 7.2x + 0.012x^2

C'(x) is the marginal cost

Revenue = x. P(x)

R(x) = x( 1700 -9x)

= 1700x - 9x^2

Differentiate R(x) with respect to x

R'(x) = 1700 - 18x

R'(x) is the marginal revenue

Profit is maximized when R'(x) = C'(x)

1700 - 18x = 500 - 7.2x + 0.012x^2

Collect like terms

0 = 500 - 1700 - 7.2x +18x +0.012x^2

0 = -1200 + 10.8x +0.012x^2

0 = 0.012x^2 + 10.8x - 1200

Using x=( -b +_ √b^2 - 4ac) /2a

a = 0.012 , b= 10.8 , c= -1200

x= (-10.8 +_ √(10.8^2) - 4*0.012*(-1200)) /2*0.012

= ( -10.8 +_√116.64 + 57.6) / 0.024

= (-10.8 +_ √174.24) / 0.024

= (-10.8 +_13.2) / 0.024

= (-10.8+13.2)/0.024 or (-10.8 - 13.2)/0.024

= 2.4/0.024 or -24/0.024

x= 100 or -1000

Since our profit cannot be negative, the profit = $100

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