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...

(The entire section contains 3 answers and 363 words.)

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