# Verifying minimum of a function when second derivative is 0

I was solving for a stable equilibrium in a game. I calculated expected payoff of one of the players which came out to be the function $$E= -4pq + 0.1p + 0.2q + 0.7$$ where p is probability of an action by one player and q is the probability of the action by opponent. Now, opponent is trying to minimize this payoff. How to go about it ? I am stuck in the maths part of the problem now.

• (I think that $p = 0.05$, not 1/2) are there any constraints on $p$ or $q$? if not, the global minimum appears to be at $(-\infty,-\infty)$ – costrom Oct 2 '15 at 21:28
• Actually I was solving for a stable equilibrium in a game. I calculated expected payoff of one of the players which came out to be the mentioned function where p is probability of an action by one player and q is the probability of the action by opponent. Now, opponent is trying to minimize this payoff. How to go about it ? I am stuck in the maths part of the problem now. – Abhinav Arya Oct 2 '15 at 21:38
• Please add the information you just mentioned to the question after marking as EDIT, so current answers become relevant. Also add the tag Game Theory if exists. – NoChance Oct 2 '15 at 21:42

HINT: Keeping $p$ constant, the given equation $E=-4pq+0.1 p+0.2 q+0.7$ represents a straight line (of form $\color{blue}{y=mx+c}$) which does not have any local minimum or maximum.
But the straight line has global minimum at $(-\infty, -\infty)$
• To the OP: to see that the equation is a straight line group the terms to become $E(q)=q(0.2-4p)+(0.7+0.1p)$. This has max and min only in a closed interval. – NoChance Oct 2 '15 at 21:40