Minimum value of the average cost

C=x3+33x+432 To find the average cost per unit,   Cxc=Cx=x3+33x+432xc=x2+33+432x Taking the first derivative of c with respect to x dcdx=2x−432x2 Taking the derivative of average cost per unit is equal to zero and we get value of x dcdx=0 ?2x−432x2=02x3−432=0 ?2x3=432x3=216x=6 unitsC=x3+33x+432 To find the average cost per unit,   Cxc=Cx=x3+33x+432xc=x2+33+432x Taking the first derivative of c with respect to x dcdx=2x−432x2 Taking the derivative of average cost per unit is equal to zero and we get value of x dcdx=0 ?2x−432x2=02x3−432=0 ?2x3=432x3=216x=6 units

a) Now evaluate the minimum value of average cost at critical number and end points of the given interval (1, 10 )

c(1)=(1)2+33+4321c(1)=466c(6)=(6)2+33+4326c(6)=141c(10)=(10)2+33+43210c(10)=176.2 The minimum value of average cost at critical number x = 6 is   141c(1)=(1)2+33+4321c(1)=466c(6)=(6)2+33+4326c(6)=141c(10)=(10)2+33+43210c(10)=176.2 The minimum value of average cost at critical number x = 6 is   141

b) Find the minimum value of average cost in given interval (10, 20 ).

We saw in the first part the function has only one critical number at x =6.

Since it is out of interval. Now evaluate the minimum value of average cost at the end points only

c(10)=(10)2+33+43210c(10)=176.2c(20)=(20)2+33+43220c(20)=454.6 The minimum value of average cost for the given interval at x= 10 is   176.2c(10)=(10)2+33+43210c(10)=176.2c(20)=(20)2+33+43220c(20)=454.6 The minimum value of average cost for the given interval at x= 10 is   176.2