I am currently trying to maximize an objective function $f(a,b,c,d,e)$ over the variable $b$ only.
By taking the derviative of f over b, setting it to zero, I can solve b in terms of the other 4 variables. So, $b=g(a,c,d,e) $
Graphically, if I substitude in some values of a,c,d,e into $f$ and using $g(a,c,d,e) $ from above to get $b$, I am able to see that $f$ reach its maximum value at that specific $b$. If I change the value of $b$, I can see that f decreases.
But then the question is, how do I prove that $b=g(a,c,d,e)$ is the point that maximize $f$? More specifically, I am looking for global maximum inside the interval [0,1].
PS: the constraint on b is that $b$ is inside the interval $[0,1]$
I find that there is one critical point of b inside [0,1].
I am not sure whether single variable calculus applies here. But if it does, since $b$ is inside the interval [0,1], and there is only one critical point (partial derivative of b is zero) inside this interval, then if $f$ doesn't diverge, the global maximum is either on the boundary or the critical point.
Do you guys think this is right?
Some said using the second derivative method. But the question is, I don't have value of other 4 variables, how do I know if the second derivative is greater than 1 or less than that?