13 18 pts The cost in dollars for a company to produce x wid

13. (18 pts) The cost, in dollars, for a company to produce x widgets is given by C(x) = 5250 + 7.00x for x greater than or equal to 0, and the price-demand function, in dollars per widget, is p(x) = 45 -0.02x for 0 less than or equal to x less than or equal to 2250.

In Quiz 2, problem #10, we saw that the profit function for this scenario is P(x) = - 0.02x^2 + 38.00x - 5250.

(c) State the maximum profit and the number of widgets which yield that maximum profit: The maximum profit is _______________ when ____________ widgets are produced and sold.

(d) Determine the price to charge per widget in order to maximize profit.

(e) Find and interpret the break-even points. Show algebraic work.

Solution

( c ) The profit function is P(x) = - 0.02x² + 38.00x – 5250. P(x) will be maximum when dP/dx= 0 and   d²P/dx² is negative. Here, dP/dx = -0.04x +38 so that if dP/dx = 0, then 0.04x = 38 or, x = 38/0.04 = 950. Also d²P/dx² = -0.04 which is negative. Hence, P(x) i.e. the profit will be maximum when x = 950 . The number of widgets that yield the maximum profit is 950. The maximum profit is -0.02(950)² + 38*950 -5250 = -18050+36100- 5250 = 12800.

( d) The price to charge per widget in order to maximize profit is p(950) = 45 -0.02 *950 = 45 – 19 = 26.

( e) The break-even point comes when the profit i.e. P(x) = 0 i.e. - 0.02x² + 38.00x – 5250 = 0 or, 2x² -3800x +52500 = 0 or, x² -1900x + 26250 = 0. On using the quadratic formula, we get x = [-(-1900)±{ (-1900)² -4*1*26250}]/2*1 = [1900± ( 810000-105000)]/2 = (1900± 705000)/2 = (1900± 839.64)/2 ( approx.). Since the maximum profit is at x = 950, for break-even, we should have x = (1900-839.64)/2 = 530.18 or, 530 ( on rounding off to the nearest whole number).

Note:

The break-even quantity (x) cannot be more than the quantity of widgets for maximum profit.