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TC = a + bq + cq²+ dq³

In the long run, none of the inputs are fixed. Thus, the fixed cost, a, becomes zero. Therefore the total cost in the long run is:

TC = bq + cq²+ dq³

The average cost function is as follows:

AC = b + cq + dq²

The quadratic equation for the AC curve shows that the average cost is either in a U-shape or inverted U-shape. A U-shape AC curve will give minimum average cost, while the inverted AC curve will cost maximum.

A firm always prefers minimum average costs. Thus, the AC curve is U-shaped.

For minimum AC, the first derivative should equal zero, which gives the critical points for q, and the second derivative of AC should be positive. Thus, at minima:

$$\begin{gathered} \frac{d\left(AC\right)}{dq}=0 \\ \frac{d\left(b+cq+d{q}^2\right)}{dq}=0 \\ c+2dq=0 \\ c=-2dq \end{gathered}$$

Thus, the average cost is minimum when c = -2dq (critical point).

For minimum average cost:

$$\begin{gathered} \frac{d^2\left(AC\right)}{d{q}^2}>0 \\ \frac{d\left(c+2dq\right)}{dq}>0 \\ 2d>0 \\ d>0 \end{gathered}$$

Since d is positive and q cannot be negative, then c has to be negative.

$$\begin{gathered} c=-2dq,or \\ q=-\frac{c}{2d} \end{gathered}$$

Placing the value of q in the AC equation:

$$\begin{gathered} AC=b+cq+d{q}^2 \\ AC=b+c\left(-\frac{c}{2d}\right)+d{\left(-\frac{c}{2d}\right)}^2 \\ AC=b-\frac{c^2}{2d}+\frac{c^2}{4d} \\ AC=b-\frac{c^2}{4d} \end{gathered}$$

The average cost must be positive, that is, AC > 0.

$Therefore,b>\frac{c^2}{4d}$

c²is positive, and d is positive. Thus, b is also positive.

Also c²< 4bd.

The marginal cost function is:

$$\begin{gathered} MC=\frac{d\left(TC\right)}{dq} \\ =\frac{d\left(bq+c{q}^2+d{q}^3\right)}{dq} \\ =b+2cq+3d{q}^2 \end{gathered}$$

The quadratic MC equation shows that the marginal cost curve is also either U-shaped or inverted U-shaped.

The marginal cost will be minimum with a U-shape curve.

For minima, the first derivate of MC equals zero will give the critical points of output.

$$\begin{gathered} \frac{d\left(b+2cq+3d{q}^2\right)}{dq}=0 \\ 2c+6dq=0 \\ c+3dq=0 \\ c=-3dq,or \\ q=-\frac{c}{3d} \end{gathered}$$

Placing the value of q in MC equation:

$$\begin{gathered} MC=b+2cq+3d{q}^2 \\ =b+2c\left(-\frac{c}{3d}\right)+3d{\left(-\frac{c}{3d}\right)}^2 \\ =b-\frac{2c^2}{3d}+\frac{c^2}{3d} \\ =b-\frac{c^2}{3d} \end{gathered}$$

MC must be positive. It implies that MC > 0. Thus,

$$\begin{gathered} b-\frac{c^2}{3d}>0 \\ b>\frac{c^2}{3d} \\ 3bd>c^2 \end{gathered}$$

From AC curve minimization analysis, it is discovered that c²< 4bd. Thus, the more rigid requirement is c²< 4bd.

Therefore, the total cost function is consistent with the U-shaped AC curve for a =0, b > 0, d > 0 and 0 < c²< 3bd.