# Thread: Absolute Extrema (two variables)

1. ## Absolute Extrema (two variables)

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

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

$\displaystyle G_{x}= y$
$\displaystyle G_{y}= x$

Therefore:

$\displaystyle y=0$ and $\displaystyle x=0$

$\displaystyle G(0,0) = 0$

Where do I go from here and how?

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

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

$\displaystyle G_{x}= y$
$\displaystyle G_{y}= x$

Therefore:

$\displaystyle y=0$ and $\displaystyle x=0$

$\displaystyle G(0,0) = 0$

Where do I go from here and how?
now find the extrema on the boundary $\displaystyle 4x^2 + y^2 = 2$

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

you can then find the absolute extrema after finding all extreme points

3. By the Extreme Value Theorem, the extremes of $\displaystyle G(x, y) = xy$ given the constraint $\displaystyle 4x^2 + y^2 \leq 2$ are either at points where $\displaystyle G_x = G_y = 0$ or on the border of the region (i.e. where $\displaystyle 4x^2 + y^2 = 2$, since this is an ellipse). Let $\displaystyle H(x, y) = 4x^2 + y^2$. Using the method of Lagrange Multipliers,

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

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

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

$\displaystyle 2y^2 = 8x^2$

$\displaystyle y = \pm 2x$

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

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

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

$\displaystyle y = \pm 1$

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