For the cost function given by C(x)=3x2+4x+243C(x)=3x2+4x+243 a

a)

The average cost is found by dividing the total cost by the number of units produced. Therefore, its function can be found by dividing C(x) by x.

¯C=C(x)x=3x2+4x+243x=3x+4+243x−1C¯=C(x)x=3x2+4x+243x=3x+4+243x−1

b)

As with any other function, we can find the critical point of the average cost function by equating its to zero and solving for x. We then check the nature of the critical point by using the second derivative test.

¯C′(x)=ddx(3x+4+243x−1)=3−243x−2¯C′(x)=0⇒3=243x−2x=±9¯C"(x)=486x−3C"(9)=0.667>0Therefore, the minimum point is x=9C¯′(x)=ddx(3x+4+243x−1)=3−243x−2C¯′(x)=0⇒3=243x−2x=±9C¯"(x)=486x−3C"(9)=0.667>0Therefore, the minimum point is x=9

Thus, the average cost is minimum when 9 units are produced.

c)

The average cost function has been graphed below.