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

2500

Explanation:

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

Monthly total revenue, R(x) =
`350x -(1)/(100)x^2` dollars

Profit = Revenue - Cost

`=R(x)-C(x)\\=(350x -(1)/(100)x^2)-(400 + 200x)\\=350x -(1)/(100)x^2-400 - 200x\\P(x)=150x-(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-(1)/(100)x^2-400\\P'(x)=150-(x)/(50)\\\\$Set $ P'(x)=0\\150-(x)/(50)=0\\150=(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)=-(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.