Fill This Form To Receive Instant Help
Homework answers / question archive / 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
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.
(b) The demand equation for a product is p=100−0.5xp=100−0.5x. The Cost function is given by C(x)=2x2+5x+18C(x)=2x2+5x+18. Find the number of items that produce maximum profit.
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.