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

Mathematics
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  • phi
you last equation does not look correct
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Other answers:

  • 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...
  • phi
btw, they are all set = to 0, and you should show this...
would that give the extreme points?
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  • phi
I think you missed a factor of 9 on the y^2 term
  • phi
but the crucial step is in solving, you should have found y=0 as a possible root
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  • phi
no. you can do with z= y^2 4z^2 +9z-9=0
ok, whats the next step?
  • 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
ok, thanks so much
  • 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

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