Suppose that a manufacturer can produce a part for $100/ unit with a fixed cost of $50,000. The alternative is to outsource production to a supplier at a unit cost of $200 Sales =−5× Price +500 Please identify the potential maximized total revenue for x-coordinate of the vertex =−b/2⋅a for a quadratic funct f(x)=ax^2+bx+c The potential maximized total revenue for the firm is: Select one: a. 25000 b. None of them C. 12500 d. 60000 e. 15650 f. 350000

The maximum total revenue, which is determined by finding the 'x' coordinate(vertex) of the given quadratic revenue function, is $12500, given the function parameters.

Explanation:

The subject of this question pertains to the maximization of revenue, which is a concept in Mathematics. Here we are looking to find the maximum point (vertex) of a quadratic equation, f(x) = ax²+bx+c, representing a revenue function, via the formula -b/2a.

From the given question, it appears that the quadratic function is the sales function, Sales=-5Price + 500. Comparing it with ax²+bx+c, we know that a=-5 (the coefficient of Price), and b=500. Substituting these values into -b/2a formula, we get -500 / (2 * -5) = 50. This is the price that maximizes the manufacturer's sales which, in turn, maximizes revenue.

So, the total maximum revenue is at the price 50 (the 'x' coordinate of the vertex of the function), therefore the revenue at that point is: R = Price * quantity = 50 * (-5*50 +500) = 12500. So, the correct choice among the options given is c. 12500.

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