Question

A firm can produce only 2500 units per month. The monthly total cost is given by C(x) = 400 + 200x dollars, where x is the number produced. If the total revenue is given by R(x) = 350x − 1 100 x2 dollars, how many items, x, should the firm produce for maximum profit?

Answer 1

Answer:

2500

Step-by-step explanation:

Monthly total cost, C(x) = 400 + 200x dollars

Monthly total revenue, R(x) = $350x -\dfrac{1}{100}x^2$ dollars

Profit = Revenue - Cost

$=R(x)-C(x)\\=(350x -\dfrac{1}{100}x^2)-(400 + 200x)\\=350x -\dfrac{1}{100}x^2-400 - 200x\\P(x)=150x-\dfrac{1}{100}x^2-400$

To determine how many items, x, the firm should produce for maximum profit, we maximize P(x) by taking its derivative and solving for its critical points.

$P(x)=150x-\dfrac{1}{100}x^2-400\\P'(x)=150-\dfrac{x}{50}\\\\ Set P'(x)=0\\150-\dfrac{x}{50}=0\\150=\dfrac{x}{50}\\ Cross multiply\\x=150*50\\x=7500$

Next, we check if the point x=7500 is a maxima or a minima.

To do this, we find the second derivative of P(x).

$P''(x)=-\dfrac{1}{50} which is negative$

Hence, the point x=7500 is a point of maxima. However, since the firm can only produce 2500 units per month.

Therefore, the company needs to produce 2500 units to maximize profit.