Why does the following not have a max/min?

Question

Let $f$ and $g$ be functions on $\mathbb{R}^{2}$ 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 $$C=\{(x,y)\in\mathbb{R}\,:\,g(x,y)=0\}.$$

Options:

• $f$ has a maximum at $(1,1)$ and minimum at $(2,2)$

• $f$ has a maximum at $(1,1)$, but does not have a minimum

• $f$ has a minimum at $(2,2)$, but does not have a maximum

• $f$ has neither a maximum nor a minimum

I got a max at $(1,1)$ and min at $(2,2)$, by taking $g(x,y)=0\Rightarrow x=y$ and thus differentiating: $$\begin{align*} \underset{x,y\in\mathcal{R}}{\max(\min)} \frac{1}{3}x^3-\frac{3}{2}y^2+2x \ & \ st.\ x-y=0\\ \Leftrightarrow \underset{x\in\mathcal{R}}{\max(\min)} \frac{1}{3}x^3-\frac{3}{2}x^2+2x \end{align*}$$

but the correct answer is:

$f$ has neither a maximum nor a minimum.

How so?

Answer 1

You found a local maximum and a local minimum. There is no global maximum because for $x=y \rightarrow \pm\infty$ the objective function $f(x,y) \rightarrow \pm \infty$.

You may also spot this by taking the limits of $\frac{1}{3}x^3-\frac{3}{2}x^2+2x$ as $x \rightarrow \pm \infty$.

Answer 2

I guess you obtained the conclusion by letting $$\left(\frac 13 x^3-\frac 32 x^2+2x\right)'=0,$$ which yields that $$(x-2)(x-1)=0,$$ and so you concluded that the function achieves extrema at $(1,1)$ and $(2,2)$, respectively. This is, however, not true, because the first-order condition only characterizes local extrema rather than global ones, or in other words, it is the necessary condition for maxima or minima, but not sufficient. In your question, it is obvious that when $x\to +\infty$, the cubic term is dominating and hence the value diverges to $+\infty$, and when $x\to -\infty$, the cubic term is also dominating and hence the value diverges to $-\infty$.