Consider a version of the Tragedy of the Commons in which herder 1 and 2 simultaneously choose to graze quantities of sheep $q _ { 1 }$ and $q 2$ ,respectively,and in which the payoff functions are $q _ { 1 } \left( 100 - q _ { 1 } - q _ { 2 } \right)$ for 1 and $q _ { 2 } \left( 100 - q _ { 1 } - q _ { 2 } \right)$ for 2.The Nash equilibrium is $q _ { 1 } ^ { * } = q _ { 2 } ^ { * } =$
A)100.
B)60.
C)40.
D)33.3.

Question

Quizplus · Accepted Answer

To find the Nash equilibrium, we must find the point where neither player can benefit by changing their strategy. This occurs where the partial derivative of each player's payoff function with respect to their strategy equals zero. 
Taking the partial derivative of player 1's payoff function with respect to $q_1$:
$$\frac{\partial}{\partial q_1} (q_1(100-q_1-q_2))=100-2q_1-q_2=0$$
Taking the partial derivative of player 2's payoff function with respect to $q_2$:
$$\frac{\partial}{\partial q_2} (q_2(100-q_1-q_2))=100-q_1-2q_2=0$$
Solving these two equations simultaneously, we get:
$$q_1^*=q_2^*=33.3$$
Therefore, the best choice is (D) 33.3 for both herders to ensure a Nash equilibrium.

Consider a Version of the Tragedy of the Commons in Which