Revenue Maximization: A company sells products at a price of $50 per unit. The demand function is p = 100 - q, where p is the price and q is the quantity sold. How many units should they sell to maximize revenue?



Answer to a math question Revenue Maximization: A company sells products at a price of $50 per unit. The demand function is p = 100 - q, where p is the price and q is the quantity sold. How many units should they sell to maximize revenue?

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To maximize revenue, we need to determine the quantity of units that will generate the highest total revenue. Revenue is calculated by multiplying the price per unit by the quantity sold. In this case, the price per unit is $50 and the quantity sold is denoted by q. The revenue function R(q) can be expressed as: R(q) = p * q Substituting the given demand function p = 100 - q, we have: R(q) = (100 - q) * q R(q) = 100q - q^2 To find the quantity that maximizes revenue, we can take the derivative of the revenue function with respect to q and set it to zero. This will give us the critical point(s) where the maximum occurs. dR/dq = 100 - 2q Setting this derivative equal to zero: 100 - 2q = 0 2q = 100 q = 50 So, the critical point is q = 50. To determine if this point corresponds to a maximum, we can take the second derivative of the revenue function with respect to q. If the second derivative is negative at q = 50, it confirms that q = 50 corresponds to a maximum. Taking the second derivative: d^2R/dq^2 = -2 Since the second derivative is negative (-2), we can conclude that q = 50 corresponds to a maximum. Therefore, to maximize revenue, the company should sell 50 units.

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