Verified answer

Profit is Maximized when Marginal Revenue = Marginal Cost. Or in a more mathematical sense:

first derivative of the revenue function = first derivative of the cost function.

We know that:

Profit = Revenue - Costs

And that:

Revenue = Price * Quantity

So for this particular problem then, we can say that the Revenue function will be:

Revenue = Price * Quantity

Revenue = x * (2800 - 9x)

The Revenue function will therefore be:

Revenue Function = -9x² + 2800x

Cost Function = 11,000 + 400x - 4.2x² + .004x³

So now in order to find the production level that will maximized profit, all we have to do is set the derivate of both the revenue and cost functions equal to each other and solve for x.

R(x) = -9x² + 2800x

C(x) = 11,000 + 400x - 4.2x² + .004x³

Profit is maximized when:

R'(x) = C'(x)

-18x + 2800 = 400 - 8.4x + .012x²

.012x² + 9.6x - 2400 = 0

Solving for x will give us:

x = -1000, 200

Since it doesn't make sense to produce a negative number of goods, we can just neglect x = -1000.

Final Answer:

Profit will be maximized when the firm produces 200 units.

Profit = Revenue - Cost

Revenue = price * quantity

= p(x) * x

= [2800 − 9x] *x

= 2800x - 9x^2

so

R(x) = 2800x - 9x^2

Profit = R(x) - C(x)

= [2800x - 9x^2] - [11000 + 400x − 4.2x^2 + 0.004x^3]

= 2800x - 9x^2 - 11000 - 400x + 4.2x^2 - 0.004x^3

so

P(x) = 2800x - 9x^2 - 11000 - 400x + 4.2x^2 - 0.004x^3

dP/dx = 2800 - 18x - 400 + 8.4x - .012x^2

so

dP/dx = - 0.012x^2 - 9.6x + 2400

set this equal to zero

- 0.012x^2 - 9.6x + 2400 = 0

you get x = -1000 or x = 200

reject the negative value

result: production level of 200 units maximizes profit.

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