To maximize profit, a firm should produce at the level where Marginal Cost (MC) equals Marginal Revenue (MR). The profit is maximized when this condition holds, provided the price is greater than the average variable cost (which is implied here as we seek maximization).
Marginal Cost is the derivative of the Cost Function with respect to quantity ($q$): $ MC = \frac{dC}{dq} = \frac{d}{dq}(5q^2) = 10q $
Marginal Revenue is the derivative of the Revenue Function with respect to quantity ($q$): $ MR = \frac{dR}{dq} = \frac{d}{dq}(50q) = 50 $
Set MC equal to MR to find the profit-maximizing quantity ($q$): $ MC = MR $ $ 10q = 50 $ $ q = \frac{50}{10} $ $ q = 5 $
The second-order condition for profit maximization requires the second derivative of the profit function to be negative. Profit $ \Pi(q) = R(q) - C(q) = 50q - 5q^2 $. $ \frac{d\Pi}{dq} = 50 - 10q $ $ \frac{d^2\Pi}{dq^2} = -10 $ Since $ -10 < 0 $, the profit is indeed maximized at $ q=5 $.
Therefore, the firm should produce 5 units to maximize profit.
The maximum value attained by the function \(f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\) in the interval [1, 2] is _____.
The function ƒ(x) = 3x (x - 2) has: