Let $f$ and $g$ be functions on $R^2$ defined respectively by $$f(x,y)=\frac{1}{3}x^3+\frac{3}{2}y^2+2x$$ and $$g(x,y)=x−y$$ Consider the problems of maximizing and minimizing $f$ on the constraint set $$C=\{{(x,y)∈R^2:g(x,y)=0\}}$$ (a) $f$ has a maximum at $(x=1,y=1)$ , and a minimum at $(x=2,y=2)$.
(b) $f$ has a maximum at $(x=1,y=1)$ , but does not have a minimum.
(c) $f$ has a minimum at $(x=2,y=2)$ , but does not have a maximum.
(d) $f$ has neither a maximum nor a minimum.
My approach
I figured the constraint set is the set of all $x=y$ replaced this in $f$ and differentiated to obtain a maxima at $(x=1,y=1)$ and a minima at $(x=2,y=2)$. The answer however states there is no maximum or minimum. Where am I at fault? Does it have something to do with the order of growth of $x^3$ ?
There exist local maxima and minima points, where the derivative vanishes. It is easy to see thta such points occur at $(-2,-2)$ and $(-1,-1)$. However, the function dosent have a lower/upper bound. Clearly, fom the constraint equation, since $x=y$,
$f(x,x) = \frac{x^3}{3} + \frac{3x^2}{2} + 2x$
clearly as $x \to +\infty$, $f(x,x) \to +\infty$ and as $x\to -\infty, f(x,x) \to -\infty$. Thus, the function has no global maxima or minima(which is what the option mramt i guess).