^{2}-140x+12200)=.001((x-70)

^{2}+(12200-4900))=.001((x-70)

^{2}+7300)

^{2}C/dx

^{2}=.002>0 so the point is a minimum.

Pie King, Inc. has determine that when x pecan pies are made daily, the average cost per pie is given by C(x)= 0.001x^2-0.14x+12.20(a) what is the average cost per pie if 30 pies are made daily?(b) how many pies should be made daily in order to minimize the average cost per pie? please show all work

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Michael F. | Mathematics TutorMathematics Tutor

C(x)= 0.001x^2-0.14x+12.20

C(30)=.001×30²-.14×30+12.2=8.9

C(x) is a parabola opening upward. The minimum is at its vertex.

C(x)=.001(x^{2}-140x+12200)=.001((x-70)^{2}+(12200-4900))=.001((x-70)^{2}+7300)

In this form the vertex is at x=70 and C(70)=7.3

If you insist on calculus dC/dx=.002x-.14, which equals 0 at x=70

and d^{2}C/dx^{2}=.002>0 so the point is a minimum.

average cost of 30 pies is substitute 30 into quadratic equation.

.001(30)^{2 }- .14(30) + 12.2

.001(900)-4.2+12.2

.9-4.2+12.2 = 8.9

to find the minimum cost per pie you need to find the vertex and the cost at the vertex.

Vertex=-b/2a.

- (-.14)/2(.001)

x=70.

Sub in 70

.001(70)^{2} - .14(70) + 12.2

4.9-9.8+12.2

7.3

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