# Math Help - Lagrange Argh...

1. ## Lagrange Argh...

Looking to find and classify critical points:

$f(x,y)=x^2+y^2+20$

subject to constraint:

$3x^2+4xy+6y^2=140$

So, let $g(x,y)=3x^2+4xy+6y^2-140=0$

$f_x+\lambda g_x=0 \Rightarrow 2x+\lambda (6x+4y)$

which gives: $\lambda(6x+4y)=-2x\Rightarrow \lambda=-\frac{x}{3x+2y}$

$f_y+\lambda g_y=0 \Rightarrow 2y+\lambda (4x+12y)$

which gives: $\lambda(4x+12y)=-2y\Rightarrow \lambda=-\frac{y}{2x+6y}$

Thus:

$\frac{x}{3x+2y}=\frac{y}{2x+6y}$

And I'm stuck...

2. Originally Posted by billym
Looking to find and classify critical points:

$f(x,y)=x^2+y^2+20$

subject to constraint:

$3x^2+4xy+6y^2=140$

So, let $g(x,y)=3x^2+4xy+6y^2-140=0$

$f_x+\lambda g_x=0 \Rightarrow 2x+\lambda (6x+4y)$

which gives: $\lambda(6x+4y)=-2x\Rightarrow \lambda=-\frac{x}{3x+2y}$

$f_y+\lambda g_y=0 \Rightarrow 2y+\lambda (4x+12y)$

which gives: $\lambda(4x+12y)=-2y\Rightarrow \lambda=-\frac{y}{2x+6y}$

Thus:

$\frac{x}{3x+2y}=\frac{y}{2x+6y}$

And I'm stuck...

...then you continue! Cross multiply and get: 2x^2 + 3yx - 2y^2 = 0 . This is a quadratic in x, and solving it we get 2(x + 2y)(x - y/2) = 0 ==> input these solutions for x now in g(x,y) and find y, and then x. I thinbk there are 4 different points, two of which are (2,4) and (-2, -4)...check this, though! I could easily be wrong...

Tonio

3. ugh... I think my brain might is broken:

for $y=2x:$

$3x^2+4x(2x)+6(2x)^2=140 \Rightarrow 35x^2=140 \Rightarrow x= \pm 2$

for $y=-x/2$ :

$3x^2+4x(-x/2)+6(-x/2)^2=140 \Rightarrow x = \sqrt {56}$

?!!

4. Originally Posted by billym
ugh... I think my brain might is broken:

for $y=2x:$

$3x^2+4x(2x)+6(2x)^2=140 \Rightarrow 35x^2=140 \Rightarrow x= \pm 2$

for $y=-x/2$ :

$3x^2+4x(-x/2)+6(-x/2)^2=140 \Rightarrow x = \sqrt {56}$

?!!

Yes...any problem? Of course, should be +/- Sqrt(56), and then you must calcualte the corresponding values for y, but it's fine imo.

Tonio

5. $(2 \sqrt{14}, - \sqrt{14}), (-2 \sqrt{14}, \sqrt{14})$ ?

6. Originally Posted by billym
$(2 \sqrt{14}, - \sqrt{14}), (-2 \sqrt{14}, \sqrt{14})$ ?

Yup...and now find the other pair, when y = 2x.

Tonio