0

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 $x^3$ ?

Community
  • 1
nsus
  • 1
  • That is indeed what i did but the my answer disagrees with the solution. – nsus Mar 07 '17 at 13:31
  • If $x - y = 0 \implies y=x$, so we have to find the extrema of $\dfrac{1}{3}x^3+\dfrac{3}{2}x^2+2x$. Since there is no constraint on $x$, what is the min and max of this function? Clear? We get no global min or max, but we do have a local min and a local max. Those local min max are located at $(x, y) = (-1, -1)$ for min and $(-2,-2)$ for max. – Moo Mar 07 '17 at 13:32
  • Yes you are right and that is why i cannot understand how the answer is d) $f$ has neither a maximum or a minimum. – nsus Mar 07 '17 at 13:36
  • Because we are looking for global versus local extrema. There are no global extrema with the contraint. As $x$ gets smaller, we go to negative infinity and as it gets larger, we go to positive infinity. The problem is asking for global extrema, not local - that is why. – Moo Mar 07 '17 at 13:39
  • Right thanks... – nsus Mar 07 '17 at 13:43

1 Answers1

0

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).

Lelouch
  • 724