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spndsh

  • 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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  1. spndsh
    • 3 years ago
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    |dw:1352550338176:dw|

  2. phi
    • 3 years ago
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    you last equation does not look correct

  3. spndsh
    • 3 years ago
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    |dw:1352551028612:dw|

  4. phi
    • 3 years ago
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    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
    • 3 years ago
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    btw, they are all set = to 0, and you should show this...

  6. spndsh
    • 3 years ago
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    would that give the extreme points?

  7. spndsh
    • 3 years ago
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    |dw:1352552076296:dw|

  8. phi
    • 3 years ago
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    I think you missed a factor of 9 on the y^2 term

  9. phi
    • 3 years ago
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    but the crucial step is in solving, you should have found y=0 as a possible root

  10. spndsh
    • 3 years ago
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    |dw:1352553005461:dw|

  11. phi
    • 3 years ago
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    no. you can do with z= y^2 4z^2 +9z-9=0

  12. spndsh
    • 3 years ago
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    ok, whats the next step?

  13. phi
    • 3 years ago
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    \[ 6x^2 -2 \lambda x=0 -> \lambda= 3x\] \[ 4y^3 -2\lambda y=0 -> y=0 \text{ or } 2y^2-3x=0 \] 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
    • 3 years ago
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    ok, thanks so much

  15. phi
    • 3 years ago
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    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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