Question

03* For which one of the following functions is (-1,-1) a relative minimum?
f(x,y)=xy + 1/x + 1/y
f(x,y)=x^2 +2x
f(x,y)=xy-y^2
f(x,y)=xy-1/x-1/y​

Answer 1

If (-1, -1) is an extremum of $f$, then both partial derivatives vanish at this point.

Compute the gradients and evaluate them at the given point.

• $f(x,y)=xy+\frac1x +\frac1y$

$\nabla f = \left\langle y - \dfrac1{x^2}, x - \dfrac1{y^2}\right\rangle \implies \nabla f (-1,-1) = \langle-2,-2\rangle \neq \langle0,0,\rangle$

• $f(x,y) = x^2+2x$

$\nabla f = \langle 2x+2,0\rangle \implies \nabla f(-1,-1) = \langle0,0\rangle$

• $f(x,y)=xy-y^2$

$\nabla f = \langle y, x-2y\rangle \implies \nabla f(-1,1) = \langle-1,1\rangle \neq\langle0,0\rangle$

• $f(x,y) = xy-\frac1x-\frac1y$

$\nabla f = \left\langle y + \frac1{x^2}, x + \frac1{y^2}\right\rangle \implies \nabla f(-1,1) = \langle0,0\rangle$

The first and third functions drop out.

The second function depends only on $x$. Compute the second derivative and evaluate it at the critical point $x=-1$.

$f(x,y) = x^2+2x \implies f'(x) = 2x + 2 \implies f''(x) = 2 > 0$

This indicates a minimum when $x=-1$. In fact, since this function is independent of $y$, every point with this $x$ coordinate is a minimum. However,

$x^2 + 2x = (x + 1)^2 - 1 \ge -1$

for all $x$, so (-1, 1) and all the other points $(-1,y)$ are actually global minima.

For the fourth function, check the sign of the Hessian determinant at (-1, 1).

$H(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -2/x^3 & 1 \\ 1 & -2/y^3 \end{bmatrix} \implies \det H(-1,-1) = 3 > 0$

The second derivative with respect to $x$ is -2/(-1) = 2 > 0, so (-1, -1) is indeed a local minimum.

The correct choice is the fourth function.