# Absolute Extrema (two variables)

• Apr 13th 2009, 01:10 PM
wilcofan3
Absolute Extrema (two variables)
I need a little help with finding absolute extrema bounded by a region.

Find the absolute extrema of $G(x,y) = xy$ over the region $4x^2+y^2\leq2$.

$G_{x}= y$
$G_{y}= x$

Therefore:

$y=0$ and $x=0$

$G(0,0) = 0$

Where do I go from here and how?
• Apr 13th 2009, 01:15 PM
Jhevon
Quote:

Originally Posted by wilcofan3
I need a little help with finding absolute extrema bounded by a region.

Find the absolute extrema of $G(x,y) = xy$ over the region $4x^2+y^2\leq2$.

$G_{x}= y$
$G_{y}= x$

Therefore:

$y=0$ and $x=0$

$G(0,0) = 0$

Where do I go from here and how?

now find the extrema on the boundary $4x^2 + y^2 = 2$

solve for $y$, say, and plug that expression in $G(x,y)$, to get a new single variable function $G(x,x)$. then find its (absolute) extreme points

you can then find the absolute extrema after finding all extreme points
• Apr 13th 2009, 01:24 PM
icemanfan
By the Extreme Value Theorem, the extremes of $G(x, y) = xy$ given the constraint $4x^2 + y^2 \leq 2$ are either at points where $G_x = G_y = 0$ or on the border of the region (i.e. where $4x^2 + y^2 = 2$, since this is an ellipse). Let $H(x, y) = 4x^2 + y^2$. Using the method of Lagrange Multipliers,

$G_x = \lambda H_x$
$G_y = \lambda H_y$

$y = \lambda \cdot 8x$
$x = \lambda \cdot 2y$

$\lambda = \frac{y}{8x} = \frac{x}{2y}$

$2y^2 = 8x^2$

$y = \pm 2x$

Substituting back into $H(x, y) = 2$

we have
$4x^2 + (2x)^2 = 2$
$4x^2 + 4x^2 = 2$
$8x^2 = 2$
$4x^2 = 1$
$x^2 = \frac{1}{4}$

$x = \pm \frac{1}{2}$

$y = \pm 1$

Hence the minimum value of $G(x, y) = xy$ is $-\frac{1}{2}$
and the maximum value of $G(x, y) = xy$ is $\frac{1}{2}$.