Answer to Question #130701 in Economics of Enterprise for adil

Question

Given TC = 100 + 60Q – 12Q2 + Q3. Find
a. The equations of the TVC, AVC, and MC functions.
b. The level of output at which AVC and MC are minimum, and prove that the AVC and MC curves are U-shaped.
c. Find the AVC and MC for the level of output at which the AVC curve is minimum.

Answer 1

We are given $TC = 100+60Q-12Q^2+Q^3$

Let us assume the currency used is $ since it is not stated in the question.

a) Calculating TVC

From the TC function, $100 is not affected by output (Q), and hence FC = $100. The remaining part of the function affected by output(Q) is the TVC function.

Thus, $TVC = 60Q-12Q^2+Q3$

Calculating AVC

$AVC = \dfrac {TVC}{Q}$

$= \dfrac {60Q-12Q^2+Q^3}{Q}$

$= \dfrac {Q(60-12Q+Q^2)}{Q}$

$= 60-12Q+Q^2$

Calculating MC

$MC = \dfrac {d}{dQ}(TC)$

$= \dfrac {d}{dQ}(100 + 60Q - 12Q^2 + Q^3)$

$= 60 - 24Q + 3Q^2$

b) Firstly, both MC and AVC are functions of the second degree because, for both, the highest power of the variable Q is 2. They are therefore quadratic functions. They are in the form $f(x) = ax^2 + bx + c$

Quadratic functions are either $\cup \space or \space \cap \space shapped$

When the coefficient of the squared term (a) is > 0 the function is minimum, and when the coefficient of the squared term (a) is < 0 the function is minimum.

Now, for both the MC and AVC, the coefficients of $Q^2$ are both positive and hence MC and AVC are minimum functions. Thus, $MC \space and \space AVC \space are \space \cup-shapped$

Finding quantity when MC and AVC are minimum.

For every quadratic function of the form $f(x) = ax^2 + bx + c$ , the minimum or maximum value of f(x) occurs when $x = \dfrac {-b}{2a}$

Thus, Q when AVC is minimum is found by $Q = \dfrac {-(-12)}{2(1)}$

$= \dfrac {12}{2}$

$= 6 \space units$

Also, Q when MC is minimum is found by $Q = \dfrac {-(-24)}{2(3)}$

$= \dfrac {24}{6}$

$= 4 \space units$

c) When AVC is minimum, Q = 6 units.

Therefore,

$AVC = 60-12(6)+6^2$

$= 60 - 72 + 36$

$= \$24$

And,

$MC = 60 - 24(6) + 3(6^2)$

$= 60 - 144 + 108$

$= \$24$

Thus, AVC = MC = $24

Note: The MC curve crosses the AVC from below and when AVC is minimum. Hence, at AVC's minimum point, it is equal to MC.