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Homework answers / question archive / For the cost function C(x)=25600+500x+x2C(x)=25600+500x+x2, find A

For the cost function C(x)=25600+500x+x2C(x)=25600+500x+x2, find A

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For the cost function C(x)=25600+500x+x2C(x)=25600+500x+x2, find

A. the cost, average cost, and marginal cost at the production level 12501250

B. The production level that will minimize the average cost

C. The minimal average cost

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To find total cost, Marginal cost and Average cost per unit to be minimized

 

A) To find the total cost, average cost and marginal cost

C(x)=25600+500x+x2Substituting value of x=1250 in given function, we get the total cost of given functionC(x)=25600+500(1250)+(1250)2C(x)=25600+625000+1562500C(x)=2213100 unitsC(x)=25600+500x+x2Substituting value of x=1250 in given function, we get the total cost of given functionC(x)=25600+500(1250)+(1250)2C(x)=25600+625000+1562500C(x)=2213100 units

To find average cost

Average cost=C(x)xAverage cost=25600+500x+x2xAverage cost=25600x+500+xWe have find average cost at production level 1250, so substituting x=1250Average cost=256001250+500+1250Average cost=20.48+500+1250Average cost=1770.48 unitsAverage cost=C(x)xAverage cost=25600+500x+x2xAverage cost=25600x+500+xWe have find average cost at production level 1250, so substituting x=1250Average cost=256001250+500+1250Average cost=20.48+500+1250Average cost=1770.48 units

To find marginal cost, taking first derivative of C(x)

C′(x)=500+2x Substituting value of x=1250C′(x)=500+2(1250)C′(x)=3000 unitsC′(x)=500+2x Substituting value of x=1250C′(x)=500+2(1250)C′(x)=3000 units

B) The production level that will minimize the average cost

C(x)=25600+500x+x2 To find the average cost per unit,   C(x)xC=C(x)x=25600+500x+x2xC=25600x+500+x Taking the first derivative of c with respect to x dCdx=1−25600x2 For the average cost per unit to be minimized, take the derivative of average cost per unit and equate it to zero dCdx=0⇒1−25600x2=0x2−25600=0x2=25600x=160 unitsC(x)=25600+500x+x2 To find the average cost per unit,   C(x)xC=C(x)x=25600+500x+x2xC=25600x+500+x Taking the first derivative of c with respect to x dCdx=1−25600x2 For the average cost per unit to be minimized, take the derivative of average cost per unit and equate it to zero dCdx=0⇒1−25600x2=0x2−25600=0x2=25600x=160 units

Thus, The production level that will minimize the average cost is 160.

C) The minimal average cost

Average cost=C(x)xAverage cost=25600+500x+x2xAverage cost=25600x+500+xWe have find minimal average cost at production level 170, so substituting x=160Minimal average cost=25600160+500+160Minimal average cost=160+500+160Minimal average cost=820 units