Find all maxima and minima for the function $f(x,y) = x^2 + y^2 -xy + 2x + 2y -4$

Question

Find all maxima and minima for the function $f(x,y) = x^2 + y^2 -xy + 2x + 2y -4$.

Here's what I attempted:

I first find the first order partial derivative $f_x(x,y) = 2x-y+2$ and $f_y(x,y) = 2y-x+2$.

Then, I equated both equations to $0$ and managed to find $x = -2 \ , \ y =-2$.

So, the critical point is $(-2,-2)$. To determine the nature of the point,

I find $D$ which is, $D = (f_{xx}(x,y))(f_{yy}(x,y)) - (f_{xy}(x,y))^2$ so

$D = (2)(2) - (-1)^2 = 3 > 0$

Since $D > 0$ and $f_{xx}(x,y) > 0$, we can conclude that the point $(-2,-2)$ is the minima.

The question mentioned find all maxima and minima but I only found the minima. Am I missing something in my answer?

Answer 1

What you did is correct. You can confirm it noticing that$$f(x,y)=\frac34(x-y)^2+\frac14(x+y+4)^2-4.$$

Answer 2

No you're not missing anything. The reason there is no maximum part of the answer is because, well, it doesn't exist. Note that $f(x,y)=x^2+(2-y)x+(y^2+2y-4)$, which is quadratic in $x$ if we fix $y$. For any fixed $y$, if we were to let $x$ get arbitrary large, the value of $f(x,y)$ would also get arbitrarily large, so no maximum exists at all.

Answer 3

your function is $$f(x,y)=\begin{bmatrix} x&y\end{bmatrix}\begin{bmatrix} 1&-1/2\\-1/2&1\end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix} -2\begin{bmatrix} x&y\end{bmatrix}\begin{bmatrix}-1\\-1\end{bmatrix} -4$$ Now as you see the $2\times 2$ matrix is positive definite hence the quadratic form i.e your function in this case will have a unique minima. As far as the maxima is concerned the range of positive definite quadratic form is $[0,\infty)$ so there is no maxima.