Final answer:
To maximize profit, the monopolist calculates the profit-maximizing quantity by setting the derivative of the profit function to zero. The demand and cost functions are used to find this quantity and the corresponding price and profit, which in this case would actually be 2500 units, $75 per unit, and $25,000 in profit, indicating a possible error in the provided choices.
Step-by-step explanation:
To find the value of x that maximizes profit for a monopolist with a demand equation p(x) = 100 − 0.01x and a cost function c(x) = 50x + 10,000, we first need to establish the revenue function, which is the number of units sold (x) multiplied by the price (p(x)).
Total Revenue, R(x), is therefore x ∙ (100 − 0.01x), which simplifies to R(x) = 100x − 0.01x^2. Total Cost, C(x), is given as 50x + 10,000. Profit, π(x), is Total Revenue minus Total Cost, so π(x) = R(x) − C(x) = 100x − 0.01x^2 − (50x + 10,000).
Simplifying the profit function, we get π(x) = 50x − 0.01x^2 − 10,000. To maximize profit, we set the derivative of the profit function to zero and solve for x. The derivative of the profit function is π'(x) = 50 − 0.02x. Setting π'(x) equal to zero gives 50 = 0.02x, so x = 2500 units.
The monopolist's profit-maximizing price is p(2500) = 100 − 0.01(2500), which simplifies to $75 per unit. The total profit for this level of production is π(2500) = 50(2500) − 0.01(2500)^2 − 10,000, which calculates to $25,000. Therefore, the correct answer is B) 1,000 units, $75 per unit, $50,000 profit, which seems to be an error since based on calculations, it should be 2500 units to maximize profit, not 1,000. Revision or clarification of the values may be needed.