a) If a cost function is given by: C(x)=2x2+255x+5000C(x)=2x2+255x+5000, find the number of items for which average cost is at a minimum

We'll first find the average cost function. The average cost is the total cost divided by the number of units. Therefore,

AC=2x2+255x+5000xAC=2x+255+5000x−1AC=2x2+255x+5000xAC=2x+255+5000x−1

Now, we differentiate the AC function and equate the derivative to zero to find the minimum point.

AC′=2−5000x−2AC′=0x2=5000/2∴x=50AC′=2−5000x−2AC′=0x2=5000/2∴x=50

Thus, the average cost is minimized at 50 units.

b)

The revenue function will be:

R=p∗xR=x∗(100−0.5x)R=−0.5x2+100xThe profit function will be:Pr=R−CPr=−0.5x2+100x−(2x2+5x+18)Pr=−2.5x2+95x−18R=p∗xR=x∗(100−0.5x)R=−0.5x2+100xThe profit function will be:Pr=R−CPr=−0.5x2+100x−(2x2+5x+18)Pr=−2.5x2+95x−18

To find the profit maximizing point, we will differentiate the profit function and equate the derivative to zero.

Pr′=−5x+95Pr′=0−5x+95=0 x=19Pr′=−5x+95Pr′=0−5x+95=0 x=19

Thus, profit is maximized at 19 units.