## spndsh 3 years ago lagrange's method: f(x,y)=2x^3+y^4 g(x,y)=x^2+y^2=1

1. spndsh

|dw:1352550338176:dw|

2. phi

you last equation does not look correct

3. spndsh

|dw:1352551028612:dw|

4. phi

I would solve for lambda using the 1st eq. plug into the 2nd equation and solve for x (or y) and finally use the last equation...

5. phi

btw, they are all set = to 0, and you should show this...

6. spndsh

would that give the extreme points?

7. spndsh

|dw:1352552076296:dw|

8. phi

I think you missed a factor of 9 on the y^2 term

9. phi

but the crucial step is in solving, you should have found y=0 as a possible root

10. spndsh

|dw:1352553005461:dw|

11. phi

no. you can do with z= y^2 4z^2 +9z-9=0

12. spndsh

ok, whats the next step?

13. phi

$6x^2 -2 \lambda x=0 -> \lambda= 3x$ $4y^3 -2\lambda y=0 -> y=0 \text{ or } 2y^2-3x=0$ notice y = 0 is a possible solution $y^2= \frac{3}{2} x$ using y^2 = 3/2 x in $x^2 + y^2 -1=0 --> x^2 +\frac{3}{2}x -1 =0$ $x= \frac{1}{4}(-3 ± 5)$ when we plug in for the various candidates (1,0) is the max value x= -2 is outside the region, x=1/2 is an inflection point

14. spndsh

ok, thanks so much

15. phi

After looking at this a little more, we have possible solutions at x=0, y=±1 y=0, x=±1 x= 1/2, y= ± 3√/2 the value of f(x,y) at these points is (0,-1) 1 (0,1) 1 (-1,0) -1 (1,0) 2 (1/2, sqrt(3)/2) 13/16 (1/2, -sqrt(3)/2) 13/16 so point (1,0) is a max with value 2 point(-1,0) is a min with value -1 points (1/2,±sqrt(3)/2) is a local min points (0,±1) are local maximums