Lagrange method to find extremes

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Homework Statement:

Study the function's maximum and minimum

Relevant Equations:

All below.
1597146354939.png

ƒ(x,y) = 3x + y
x² + 2y² ≤ 1

It is easy to find the maximum, the really problem is find the minimum, here is the system:

(3,1) = λ(2x,4y)
x² + 2y² ≤ 1

how to deal with the inequality?
 

Answers and Replies

  • #2
Office_Shredder
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Sorry, I find the picture a little unclear. f(x,y)=3x+y is the function you're maximizing/minimizing, and ##x^2+2y^2\leq1## is the region of the plane you're doing it in?

The extrema all are in one of two states:
1.) The inequality is strict. In that case there's no boundary condition, and the gradient needs to just be zero.

2.) The inequality is exact. In that case you can use Lagrange multipliers to figure out where the extrema are.

I think you need to just solve these two cases separately.
 
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  • #3
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Sorry, I find the picture a little unclear. f(x,y)=3x+y is the function you're maximizing/minimizing, and ##x^2+2y^2\leq1## is the region of the plane you're doing it in?

The extrema all are in one of two states:
1.) The inequality is strict. In that case there's no boundary condition, and the gradient needs to just be zero.

2.) The inequality is exact. In that case you can use Lagrange multipliers to figure out where the extrema are.

I think you need to just solve these two cases separately.
The gradients i found can never be zero, that is:

∇ƒ = (3,1,0)
so what to do?

I though wrongly that it would be, for example, (0,0) the minimum, but we can find to (-1,0) and so f = -3.
 
  • #4
Office_Shredder
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If the gradient is never zero, then all the extrema must be where the inequality is an equality, and you can use Lagrange multipliers.
 
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  • #5
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If the gradient is never zero, then all the extrema must be where the inequality is an equality, and you can use Lagrange multipliers.
Okay, that's good to know, i found:

1597336861797.png


the book:

1597336589759.png

but (-6/√38,-6/√38) dont satisfy!
(-6/√38)² + 2*(-6/√38)² ≤ 1

Maybe am i right?
 

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  • #6
Office_Shredder
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I think you got it right, and the book has a typo.
 
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