# find the production level that will maximize profit. If  C(x) = 15000 + 600x − 2.8^2+ 0.004x^3 is the cost function and  p(x) = 4200 − 7x  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.)

The cost function is C(x)=15000+600x-2.8x^2+0.004x^3  and the revenue function is P(x)=4200-7x

Here we have to find the production level that will maximize the profit.

For maximum profit, the marginal cost should be equal to the marginal revenue.

Now, the marginal cost is the derivative of cost function.

C'(x)=600-5.6x+0.012x^2

Similarly, the marginal revenue is the derivative of the revenue function.

P'(x)=-7

When we equate both, we get,

C'(x)=P'(x)

600-5.6x+0.012x^2=-7

0.012x^2-5.6x+607=0

Now,

x=\frac{5.6\pm \sqrt{5.6^2-4(0.012)(607)}}{2(0.012)}

=\frac{5.6\pm \sqrt{2.224}}{0.024}

So x=295.46 \ or \ x=171.21

In other words, x\approx 295\ units \ or \ x=171\ units

The cost function...

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