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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352550338176:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.0you last equation does not look correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352551028612:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I 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...

phi
 3 years ago
Best ResponseYou've already chosen the best response.0btw, they are all set = to 0, and you should show this...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would that give the extreme points?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352552076296:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I think you missed a factor of 9 on the y^2 term

phi
 3 years ago
Best ResponseYou've already chosen the best response.0but the crucial step is in solving, you should have found y=0 as a possible root

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352553005461:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.0no. you can do with z= y^2 4z^2 +9z9=0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, whats the next step?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0\[ 6x^2 2 \lambda x=0 > \lambda= 3x\] \[ 4y^3 2\lambda y=0 > y=0 \text{ or } 2y^23x=0 \] 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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0After 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
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