Maximum and minimum of $\frac13x^3 - \frac32y^2 + 2x $ such that $x-y=0$

Question

The function to maximize and minimize is;

$$f(x,y) = \frac13x^3 - \frac32y^2 + 2x $$

The constraint is;

$$g(x,y) = x-y $$

such that $$g(x,y) = 0$$

I found the first order condition and found critical points at x = 2 and x = 1.

The second order condition at $x = 1$ is -ve and at $x = 2$ is +ve.

So they should be maximum and minimum respectively. But my textbook disagrees. It says that $f(x,y)$ neither has a maxima or minima. Can you please help ?

Answer 1

Observe that $x-y=0\implies x=y$. Thus $$\begin{align*}f(x,y)&=\frac13x^3 - \frac32y^2 + 2x\\&=\frac13x^3 - \frac32x^2 + 2x\\\end{align*}$$

which is a cubic equation. Thus $x\to\pm\infty\implies f\to\pm\infty$. Thus there are no global maximum or minimum...

Answer 2

The condition is $g(x,y) = 0 \Leftrightarrow x=y$.

So, the problem becomes a simple single-variable calculus problem. You want to find the maximum and minimum of $h(x) :=f(x,x) = \frac13x^3 - \frac32x^2 + 2x$.

This is a cubic polynomial, so it does not have a global maximum or a global minimum since $$\lim\limits_{x\rightarrow \infty}h(x) = \infty \text{ and } \lim\limits_{x\rightarrow - \infty}h(x) = -\infty.$$

So, your book is right.

Now, $h'(x) = x^2-3x+2$, which vanishes at $x=1$ and $x=2$. These are the local minimum and the local maximum of $h$ respectively.

Answer 3

For a problem like this, you can also use Lagrange Multipliers.

We have $g(x,y)=\frac{1}{3}x^3-\frac{3}{2}y^2+2x$ and $h(x,y)=x-y=0$.

$\frac{\partial g}{\partial x}=\lambda\frac{\partial h}{\partial x}$ implies:

$x^2+2=\lambda$

$\frac{\partial g}{\partial y}=\lambda \frac{\partial h}{\partial y}$ implies:

$-3y=-\lambda$, so $\lambda = 3y$.

Using our constraint $h(x,y)$, we have $y=x$, so $\lambda = 3x$.

Finally we have $x^2+2=3x$.

This polynomial has roots at $1$ and $2$.

Check the values of $g(x,y)$ there to sort out the values. Also check $g(x,y)$ at $\infty$ and $-\infty$. Extreme values occur at critical points and at the boundary(including the ifinities).

You get the same results as other methods, but it can make problems easier with more variables or more complicated constraints.