Answer:
2500
Step-by-step explanation:
Monthly total cost, C(x) = 400 + 200x dollars
Monthly total revenue, R(x) = [tex]350x -\dfrac{1}{100}x^2[/tex] dollars
Profit = Revenue - Cost
[tex]=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[/tex]
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.
[tex]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[/tex]
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).
[tex]P''(x)=-\dfrac{1}{50} $ which is negative[/tex]
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.
