Answer to Question #253855 in Microeconomics for Tami

Question

A firm's cost curve is C=F+10q-bq²+q³,where b>0. 1.For what value of b are cost,average cst,and average variable cost positive?( From now on, assume that all these measures of cost are postive at every output level.)

2.What is the shape of the AC curve At what output level is the AC minimized?

3.At what output levels does the MC curve cross the AC and the AVC curves?

4. Use calculus to show that the MC curve must cross the AVC at its minimum point.

Answer 1

1.

$C = F + 10q - bq^2 + q^3$

Average Cost (AC) = $\frac{TC }{ q}$

AC = $\frac{(F + 10q - bq^2 + q^3) }{ q}$

$AC = \frac{F}{q} + 10 - bq + q^2$

TVC = 10q - bq2 + q3

AVC =$\frac{ TVC }{ q}$

AVC $= \frac{(10q - bq^2 + q^3) }{ q}$

$AVC = 10 - bq + q^2$

TC is positive if

TC > 0

$F + 10q - bq^2 + q^3 > 0$

$F + 10q + q^3 > bq^2$

$b < \frac{(F + 10q + q^3)}{ q^2}$

2.

$AC = \frac{F}{q} + 10 - bq + q^2.$ It is an U-shaped curve.

AC is minimized when the first order derivate of AC with respect to q is equal to zero.

$\frac{d(AC)}{dq} = (-\frac{F}{q^2}) - b + 2q = 0$

3,

$MC = \frac{d(TC)}{dq}$

$MC = 10 - 2bq + 3q^2$

When MC crosses AC then, MC = AC

$10 - 2bq + 3q^2 = \frac{F}{q} + 10 - bq + q^2$

$-b + 2q - (\frac{F}{q^2}) = 0$

4.

$AVC = 10 - bq + q^2$

$\frac{d(AVC)}{dq} = -b + 2q$

AVC will be at minimum when $\frac{d(AVC)}{dq} = 0$

-b + 2q = 0

$q =\frac{ b}{2}$

$MC = 10 - 2bq + 3q^2$

at this MC = AVC