Example

A producer is a price-taker with cost function $C(y)=6y^3-30y^2+100y$. We determine the supply function.

The average cost function is given by $AC(y)=6y^2-30y+100$. According to the first-order condition for an extremum the average costs are at a minimum at a stationary point of $AC(y)$,
$$\begin{align} (AC'(y)=)\;12y-30=0 &\Leftrightarrow & 12y=30\\                    &\Leftrightarrow & y=2\tfrac{1}{2}. \end{align}$$

Since $AC''(y)=12$ it holds that $AC''(2\frac{1}{2})=12>0$. Hence, according to the second-order condition for an extremum $AC(2\frac{1}{2})=62\frac{1}{2}$ is the minimum of $AC(y)$.

Now we distinguish between two cases: $p<\min AC(y)=62\frac{1}{2}$ and $p\geq \min AC(y)=62\frac{1}{2}$. For $p<62\frac{1}{2}$ the producer decides, based on the production rule, not to produce, which implies $y=0$. For $p \geq 62\frac{1}{2}$ the producer uses the marginal output rule to determine the supply that results in maximum profit.

The profit function of the producer is given by
$$\pi(y)=py-(6y^3-30y^2+100y).$$
According to the first-order condition the most profitable output quantity is a stationary point of $\pi$,
$$\begin{align} \pi'(y)=0 &\Leftrightarrow& p-(18y^2-60y+100)=0\\ &\Leftrightarrow& -18y^2+60y-100+p = 0. \end{align}$$
Hence,
$$y=\frac{-60 - \sqrt{60^2 -4\cdot(-18)\cdot(-100+p)}}{-36} =1\tfrac{2}{3}+\tfrac{1}{36}\sqrt{72p-3600}$$
and
$$y=\frac{-60 + \sqrt{60^2 -4\cdot(-18)\cdot(-100+p)}}{-36} =1\tfrac{2}{3}-\tfrac{1}{36}\sqrt{72p-3600}.$$
Since $\pi'(y)$ is a quadratic function with $a<0$ we find the maximum profit for an output quantity of $y=1\frac{2}{3}+\frac{1}{36}\sqrt{72p-3600}$, whenever $p\geq 62\frac{1}{2}$. We conclude that the supply function is defined by
$$y(p)=\left \{ \begin{array}{ll} 0                          & {\rm if } \ p<62\frac{1}{2}\\ 1\frac{2}{3}+\frac{1}{36}\sqrt{72p-3600}       & {\rm if } \ p\geq 62\frac{1}{2}. \end{array} \right.$$