The cost function for a telephone manufacturing plant is... Determine the following...

Question

Question #1:

The cost function for a telephone manufacturing plant is  and the revenue function is , where x is the number of telephones sold, in thousands.

a) Determine the profit function for the number of phones sold.

b) Determine the number of telephones sold that maximizes profit.

Please make your answer easy to understand...thank you. :)

Answer 1

Profit = Revenue minus Cost

P(x) = R(x) - C(x) = -2x^2 + 27x - (5x+24)

= -2x^2 +27x -5x -24

= -2x^2 +22x -24

P(x) = -2x^2 +22x -24

take the derivative and set equal to zero

P'(x) = -4x + 22x - = 0

divide by -2

2x -11x = 0

x = 11/2 = 5.5 thousand = 5,500 phones maximizes profit

P(x) = -2x^2 + 22x -24

=-2(x^2 -11x ) - 24

= -2(x^2 -11x + (11/2)^2) - 24 + (11/2)^2

= -2(x- 5.5)^2 - 25/4

= -2(x- 5.5)^2 - 6.25

vertex = maximum = (5.5, -6.25)

maximum "profit" = minimum loss = -6.25

there is no non-negative output level witha positive profit.

best to file bankruptcy, or sell the business

or some less ethical owners abscond with all the assets they can steal and disappear, pretending

the assets and cash were just part of their salary.

the higher x, after x = 5.5, the bigger the loss

P(x) = -2x^2 + 22x -24 is a downward opening parabola

=-2(x^2 -11x ) -24

= -2(x^2 - 11x + (11/2)^2) -24 + (11/2)^2

=-2(x- 11/2)^2 -25/4

with vertex = maximum point = (11/2, -25/4) = (5.5, -6.25) = (x,P) with "profit" maximizing output = 5.5 and maximum "profit "= negative 6.25