# Math Help - the last part

1. ## the last part

find the extremea of f(x,y) = x^2 -4y^2 subject to g(x,y) = x^2+ 4xy +6y^2 =2o.

making use of lagrange multiplier, i got

f(x,y) has a min or max when x^2 +5xy+4y^2 = o and that it has to also fulfil the conditions of g(x,y) = x^2+ 4xy +6y^2 =2o right?

when i solve these 2 equations via simultaneous, i got y=0 and x=2y. then how do i solve the last part?

2. Originally Posted by alexandrabel90
find the extremea of f(x,y) = x^2 -4y^2 subject to g(x,y) = x^2+ 4xy +6y^2 =2o.

making use of lagrange multiplier, i got

f(x,y) has a min or max when x^2 +5xy+4y^2 = o and that it has to also fulfil the conditions of g(x,y) = x^2+ 4xy +6y^2 =2o right?

when i solve these 2 equations via simultaneous, i got y=0 and x=2y. then how do i solve the last part?
It took me a moment to see how you got $x^2+ 5xy+ 4y^2= 0$ but, yes, that is correct. Subtracting that from $x^2+ 4xy+ 6y^2= 20$ gives $-xy+ 2y^2= 20$. That does NOT give "y= 0 and x= 2y". They satisfy $y(x- 2y)= y^2- 2y= 0$. Did you forget the "20"?

You can solve $-xy+ 2y^2= 20$ for x as $x= \frac{2y^2- 20}{y}$. Put that back into, say, $x^2+ 4xy+ 6y^2= 20$ to get an equation for y alone.

3. subbing x into x^2+ 4xy +6y^2 =2o , i got 10y^4 -172y^2 +400=0.

so y^2 = (172 + (29544)^(1/2) )/ 20 or y^2 = (172 - (29544)^(1/2) )/ 20

then how do i get an equation in terms of just y alone and not y^2?

from here, what do i sub into x and y to find the points where f has a local minima and maxima?

thanks!