Question 6.32: The total variable cost, TVC, and total fixed cost, TFC, fun......

(a)

AC=\frac{TC}{Q}=\frac{\displaystyle\frac{\displaystyle\mathrm{Q}^{3}}{2}-15{Q}^{2}+175{Q}+1000}{Q} =\frac{Q^{2}}{2}  –  15Q  +  175  + \frac{1000}{Q}

AVC = \frac {TVC}{Q} = \frac {\frac {Q³}{2}− 15Q² + 175Q }{Q} = \frac {Q²}{2} − 15Q + 175

AFC = \frac {TFC}{Q} = \frac {1000}{Q}

MC = \frac {d(TC)}{dQ} = \frac {3Q²}{2} − 30Q + 175 or MC = \frac {d(TVC)}{dQ} = \frac {3Q²}{2} − 30Q + 175

Notice that AC = AVC + AFC.

(b) The turning points for MC, AVC, AFC and AC are now calculated.

• To find the minimum values for the MC and AVC functions equate the first derivatives to zero. The second derivative should be positive at each minimum point.

MC = \frac {3Q²}{2} − 30Q + 175

Step 1: Find first and second derivatives

\frac {d(MC)}{dQ} = \frac {3(2Q)}{2} − 30(1) = 3Q − 30

\frac {d²(MC)}{dQ²} = 3

Step 2: Find turning points by solving first derivative = 0.

3Q − 30 = 0 → Q = 10

Step 2a: When Q = 10, MC = 25

Step 3: Confirm the minimum.

The second derivative is a positive constant, so the only turning point is a minimum.

AVC = \frac {Q²}{2} − 15Q + 175

Step 1: Find first and second derivatives

\frac {d(AVC)}{dQ} = \frac {(2Q)}{2} − 15(1) = Q − 15

\frac {d²(AVC)}{dQ²} = 1

Step 2: Find turning points by solving first derivative = 0.

Q − 15 = 0 → Q = 15

Step 2a: When Q = 15, AVC = 62.5

Step 3: Confirm the minimum.

The second derivative is a positive constant, so the only turning point is a minimum.

The graphs of the AVC and MC functions are plotted in Figure 6.38(b) from the points in columns (1), (6) and (8) of Table 6.18.

• The graph of AFC = 1000/Q is a hyperbolic curve with the vertical axis as asymptote.

Step 1: Get derivatives:

AFC = 1000Q^{−1}

Step 2: At a minimum, the first derivative is zero. Solve this equation for Q:

– \frac {1000}{Q²} =0  → −1000 = 0(Q²) = 0

This statement, −1000 = 0, is a contradiction, therefore no solution. To put it another way, there is no value of Q² which can be multiplied by 0 to give −1000. Therefore, since we cannot find a value of Q for which slope is zero, there is no turning point.

Curvature: Slope

\frac {d(AFC)}{dQ} = – \frac {1000}{Q²}

is negative for all values of Q, therefore this is a decreasing curve. The second derivative,

\frac {d^2(AFC)}{dQ^2} = \frac {2000}{Q³}

is positive for all values of Q > 0, therefore the curvature is concave up or convex towards the origin.

• The point where AC is at a minimum is not so easily calculated. Equating the derivative of AC to zero and solving for Q produces a cubic equation, as shown below:

AC = 0.5Q² − 15Q + 175 + 1000Q^{−1}

\frac {d(AC)}{dQ} = 0.5(2Q) − 15(1) + 1000(−Q^{−2}) = Q − 15 − 1000Q^{−2}

At the minimum AC, the first derivative is zero,

Q − 15 + \frac {1000}{Q²} = 0

\frac {Q}{1}  –  \frac {15}{1}  +  \frac {1000}{Q²} =0

\frac {Q³  –  15Q²  +  1000}{Q²} =0

Q³ − 15Q² + 1000 = 0         multiplying both sides by Q²

The solution to cubic equations is beyond the scope of this text, unless it is possible to factor the cubic function, thereby reducing the problem to the product of linear and quadratic functions. However, you are expected to be resourceful, so when all else fails, sketch the graph by plotting the points from columns (1) and (7) in Table 6.18, and estimate from the graph that minimum AC is approximately at Q = 18.1 and AC = 123. In this case, it is quite safe to plot points, since polynomials, like this one, have no infinite jumps, etc. The AC function is plotted in Figure 6.38(b).
Since the minima for the MC and AVC are at Q = 10 and Q = 15, respectively, sketch the graphs from Q = 0 to Q = 30 to get a reasonable picture of the main points of all curves. Use Excel, if available, to calculate the tables of points and plot the graphs.

(c) From Figure 6.38(b), we can deduce the following:

• The MC curve passes through the minimum point of the AVC curve (MC = AVC when AVC is at a minimum).
  The minimum value of AVC is at Q = 15, AVC = 62.5. To show that MC and AVC curves intersect at this point, evaluate MC at Q = 15:

MC = 1.5Q² − 30Q + 175
= 1.5(15)² − 30(15) + 175
= 337.5 − 450 + 175 = 62.5

So both curves pass through the point Q = 15, MC = AVC = 62.5. This confirms that
MC = AVC at minimum AVC.

• The MC curve passes through the minimum point of the AC curve (MC = AC when AC is a minimum).
  AC is at a minimum at Q = 18.1 and AC = 123. To show that MC and AC curves intersect at the point Q = 18.1, evaluate MC at Q = 18.1:

MC = 1.5Q² − 30Q + 175
= 1.5(18.1)² − 30(18.1) + 175
= 491 − 543 + 175 = 123

This confirms that MC = AC when AC is a minimum. To summarise:

When AC is at a minimum, MC = AC.
When AC is falling, MC < AC; when AC is rising, MC > AC.
When AVC is at a minimum, MC = AVC.
Minimum MC < Minimum AVC < Minimum AC.