Exercise 2 | Wiskunde op Tilburg University

A producer is a price-taker with cost function

$C(y)=10y^3-60y^2+90y$ . Determine the supply function.

Antwoord 1 correct

Correct

Antwoord 2 optie

$y(p)=\left \{ \begin{array}{ll} 0 & {\rm if } \ p<3\\ 2+\frac{1}{60}\sqrt{120p+3600} & {\rm if } \ p\geq 3 \end{array} \right.$

Antwoord 2 correct

Fout

Antwoord 3 optie

$y(p)=2+\tfrac{1}{60}\sqrt{120p+14310}$

Antwoord 3 correct

Fout

Antwoord 4 optie

$y(p)=\left \{ \begin{array}{ll} 0 & {\rm if } \ p<3\\ 2+\frac{1}{60}\sqrt{120p+14310} & {\rm if } \ p\geq 3 \end{array} \right.$

Antwoord 4 correct

Fout

Antwoord 1 optie

$y(p)=2+\tfrac{1}{60}\sqrt{120p+3600}$

Antwoord 1 feedback

Correct:

$AC(y)=10y^2-60y+90$ Then

$AC'(y)=20y-60$ gives

$y=3$ . Since

$AC''(y)=20$ it holds that

$AC''(3)=20>0$ . Hence, according to the second-order condition for an extremum

$AC(3)=0$ is the minimum of

$AC(y)$ .

The profit function of the producer is given by

$\pi(y)=py-(10y^3-60y^2+90y)$ . which gives

$\begin{align*} \pi'(y)=0 &\Leftrightarrow& p-(30y^2-120y+90)=0\\ &\Leftrightarrow& -30y^2+120y-90+p = 0. \end{align*}$
Hence,

$y=\frac{-120 - \sqrt{120^2 -4\cdot(-30)\cdot(-90+p)}}{-36} =2+\sqrt{120p+3600}$
and

$y=\frac{-120 + \sqrt{120^2 -4\cdot(-30)\cdot(-90+p)}}{-36} =2-\sqrt{120p+3600}.$
Since

$\pi'(y)$ is a quadratic function with

$a<0$ we find the maximum profit for an output quantity of

$y=2+\sqrt{120p+3600}$ , whenever

$p\geq 0$ . We conclude that the supply function is defined by

$y(p)=2+\frac{1}{60}\sqrt{120p+3600}$ .

Go on.

Antwoord 2 feedback

Wrong:

$y=3$ is the minimum location of the average cost function, not the minimum itself.

See Minimum/maximum.

Antwoord 3 feedback

Wrong: Consider minus-signs when working with brackets.

Try again.

Antwoord 4 feedback

Wrong:

$y=3$ is the minimum location of the average cost function, not the minimum itself.

See Minimum/maximum.