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
\[
y(p)=\left \{
\begin{array}{ll}
0 & {\rm if } \ p<3\\
2+\frac{1}{60}\sqrt{120p+3600} & {\rm if } \ p\geq 3
\end{array}
\right.
\]
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
\[
y(p)=\left \{
\begin{array}{ll}
0 & {\rm if } \ p<3\\
2+\frac{1}{60}\sqrt{120p+14310} & {\rm if } \ p\geq 3
\end{array}
\right.
\]
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
\[
y(p)=2+\tfrac{1}{60}\sqrt{120p+3600}
\]
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.
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.
See Minimum/maximum.
Antwoord 3 feedback
Wrong: Consider minus-signs when working with brackets.
Try again.
Try again.
Antwoord 4 feedback
Wrong: $y=3$ is the minimum location of the average cost function, not the minimum itself.
See Minimum/maximum.
See Minimum/maximum.