Chapter 6, Problem 14AP

a)

To determine

The supply curve derivation for hospitals 1 and 2.

Explanation of Solution

Consider the profit of hospital over the two-year period as follows:

$\text{Profit}(\prod )\text{=}\left({\text{P}}_{\text{1}}-{\text{C}}_{\text{1}}\right){\text{q}}_{\text{1}}\text{+}\left({\text{P}}_{\text{2}}-{\text{C}}_{\text{2}}\right){\text{q}}_{\text{2}}$

Where,

$\begin{array}{l}{\text{P}}_{\text{1}},{\text{P}}_{\text{2}}=\text{Constant}\\ {\text{C}}_{\text{1}}=\frac{{\text{q}}_1}{2}\\ {\text{C}}_{\text{2}}=\frac{{\text{q}}_2}{2}\end{array}$

Then, rewrite the profit equation as follows:

$\begin{array}{c}\text{Profit}(\prod )\text{=}\left({\text{P}}_{\text{1}}-\frac{{\text{q}}_{\text{1}}}{2}\right){\text{q}}_{\text{1}}\text{+}\left({\text{P}}_{\text{2}}-\frac{{\text{q}}_{\text{2}}}{2}\right){\text{q}}_{\text{2}}\\ ={\text{P}}_{\text{1}}{\text{q}}_{\text{1}}+{\text{P}}_{\text{2}}{\text{q}}_{\text{2}}-{\text{q}}_{\text{1}}{}^{\frac{2}{2}}-{\text{q}}_{\text{2}}{}^{\frac{2}{2}}\end{array}$

Differentiate q1 by keeping q2 as constant to obtain as follows:

$\frac{\text{dx}}{\text{dq}}\text{=}0{\text{+P}}_{\text{2}}-{\text{q}}_{\text{2}}$

Differentiate q2 by keeping q1 as constant to obtain as follows:

$\frac{\text{dx}}{{\text{dq}}_2}\text{=}0{\text{+P}}_{\text{1}}-{\text{q}}_{\text{1}}$

The derivative of profit will be zero for profit maximizing. Then, equate the supply function as follows:

$\begin{array}{l}{\text{P}}_{\text{1}}-{\text{q}}_{\text{1}}=0\\ \therefore {\text{P}}_{\text{1}}={\text{q}}_{\text{1}}\end{array}$

b)

To determine

The supply curve derivation for hospitals 1 and 2.

Explanation of Solution

Consider the profit of hospital over the two-year period as follows:

$\text{Profit}(\prod )\text{=}\left({\text{P}}_{\text{1}}-{\text{C}}_{\text{1}}\right){\text{q}}_{\text{1}}\text{+}\left({\text{P}}_{\text{2}}-{\text{C}}_{\text{2}}\right){\text{q}}_{\text{2}}$

Where,

$\begin{array}{l}{\text{P}}_{\text{1}},{\text{P}}_{\text{2}}=\text{Constant}\\ {\text{C}}_{\text{1}}=\frac{{\text{q}}_1}{2}\\ {\text{C}}_{\text{2}}=\frac{{\text{q}}_2}{2{\text{q}}_1}\end{array}$

Then, rewrite the profit equation as follows:

$\begin{array}{c}\text{Profit}(\prod )\text{=}\left({\text{P}}_{\text{1}}-\frac{{\text{q}}_{\text{1}}}{2}\right){\text{q}}_{\text{1}}\text{+}\left({\text{P}}_{\text{2}}-\frac{{\text{q}}_{\text{2}}}{2{\text{q}}_{\text{1}}}\right){\text{q}}_{\text{2}}\\ ={\text{P}}_{\text{1}}{\text{q}}_{\text{1}}+{\text{P}}_{\text{2}}{\text{q}}_{\text{2}}-{\text{q}}_{\text{1}}{}^{\frac{2}{2}}-{\text{q}}_{\text{2}}{}^{\frac{2}{2{\text{q}}_{\text{1}}}}\end{array}$

Differentiate q2 by keeping q1 as constant to obtain as follows:

$\begin{array}{l}\frac{\text{dx}}{{\text{dq}}_2}\text{=}0{\text{+P}}_{\text{2}}-\frac{{\text{q}}_{\text{2}}}{{\text{q}}_{\text{1}}}\\ \text{let, }\frac{\text{dx}}{{\text{dq}}_2}\text{=}0\\ \therefore {\text{q}}_{\text{2}}={\text{P}}_{\text{2}}{\text{q}}_{\text{1}}\end{array}$

Substitute this in the equation for profit to obtain as follows:

$\begin{array}{c}\text{Profit}(\prod )\text{=}\left({\text{P}}_{\text{1}}-\frac{{\text{q}}_{\text{1}}}{2}\right){\text{q}}_{\text{1}}\text{+}\left({\text{P}}_{\text{2}}-\frac{{\text{q}}_{\text{2}}}{2{\text{q}}_{\text{1}}}\right){\text{q}}_{\text{2}}\\ ={\text{P}}_{\text{1}}{\text{q}}_{\text{1}}+{\text{P}}_{\text{2}}{}^{{\text{2q}}_{\text{1}}}-{\text{q}}_{\text{1}}{}^{\frac{2}{2}}-{\text{P}}_{\text{2}}{{}^{{\text{2q}}_{\text{1}}}}^{\frac{2}{2{\text{q}}_{\text{1}}}}\\ ={\text{P}}_{\text{1}}{\text{q}}_{\text{1}}+{\text{P}}_{\text{2}}{}^{\frac{{\text{2q}}_{\text{1}}}{2}}-{\text{q}}_{\text{1}}{}^{\frac{2}{2}}\end{array}$

$\begin{array}{l}\frac{\text{dx}}{{\text{dq}}_2}{\text{=P}}_{\text{1}}+{\text{P}}_{\text{2}}{}^{\frac{2}{2}}-{\text{q}}_1=0\\ {\text{q}}_1={\text{P}}_{\text{1}}+{\text{P}}_{\text{2}}{}^{\frac{2}{2}}\\ {\text{q}}_2={\text{P}}_{\text{1}}{\text{P}}_{\text{2}}+{\text{P}}_{\text{2}}{}^{\frac{3}{2}}\end{array}$

c)

To determine

Identify the patients rush in period 1 if there is a possibility of learning by doing.

Explanation of Solution

The hospital will see more patients at the period one if there is learning by doing as per the equation of supply.