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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. 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
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