A firm's average cost function is given by $A C = \frac { 800 } { Q } + 2 Q + 18$ Find, to the nearest whole number, the value of Q at the lowest point on the graph of AC plotted against Q, in the interval, 0 c Q c 30.
A)25
B)20
C)15
D)5
E)10

Question

Quizplus · Accepted Answer

To find the value of $Q$ at the lowest point on the graph of $AC$ against $Q$, we need to differentiate the average cost function with respect to $Q$ and set the derivative equal to zero to find the critical points. The average cost function is given by $AC = \frac{800}{Q} + 2Q + 18$.Differentiating $AC$ with respect to $Q$ gives:$$AC' = -\frac{800}{Q^2} + 2$$Setting $AC'$ equal to zero and solving for $Q$ gives:$$-\frac{800}{Q^2} + 2 = 0$$$$\Rightarrow 2 = \frac{800}{Q^2}$$$$\Rightarrow Q^2 = \frac{800}{2}$$$$\Rightarrow Q^2 = 400$$$$\Rightarrow Q = 20$$Therefore, the value of $Q$ at the lowest point on the graph of $AC$ plotted against $Q$ is 20, to the nearest whole number.

A Firm's Average Cost Function Is Given By AC=800Q+2Q+18A C = \frac { 800 } { Q } + 2 Q + 18AC=Q800​+2Q+18

A Firm's Average Cost Function Is Given By $A C = \frac { 800 } { Q } + 2 Q + 18$