^{2}- 100X + 8300

^{2}- 100(50) + 8300 = 5800

The Marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the marginal cost C (in dollars) to produce x thousand mp3 players is given by the function C(x)=x^2-100x+8300. (a). How many players should be produced to minimize the marginal cost? and (b). What is the minimum cost??

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

If you take Calculus marginal is the derivative, the value of change at a point(margin).

C( X) = X^{2} - 100X + 8300

a) This is a quadratic function, the minimum value is @ - b/ 2a:

- ( -100) /2 = 50

b)

C( 50) = (50) ^{2} - 100(50) + 8300 = 5800

Another way, using Calculus:

a) dC/dx = 2x - 100 =50

The Marginal cost is found by taking the derivative of the cost equation.

C(x) = x^2 - 100x + 8300

C'(x) = 2x - 100

This curve will have a minimum since it's a parabola that is concave up.

2x - 100 = 0

2x = 100

x = 50

50,000 mp3 players should be made to minimize the marginal cost

The minimum cost will be 50^2 - 100*50 + 8300 = $5800

Hey Kevin -- at a minimum, the slope of the trough is zero ... take dC(X)/dX =0 ...

a) 2x -100 =0 ... x= 50 ==> **produce 50k mp3 players**

b) C(50) = 50*50 less 5k plus 8300 = **5800 min cost** at x= 50 ... Best wishes, sir :)

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