(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I seem to be struggling a bit to understand how my prof solved this problem....I think it might be my diminishing system of equation skills, so forgive me if this doesn't belong in the calc section.

Use Lagrange multipliers to find all extrema of the function subject to the given constraint:

[tex]f(x,y)=y-x^2[/tex]

subject to:

[tex]g(x,y)=x^2+y^2=1[/tex]

2. Relevant equations

The local extrema should exist where the gradient of the function is equal to the gradient of the constraint, multiplied by a value (the Lagrange multiplier):

[tex]\nabla f = \lambda\nabla g[/tex]

3. The attempt at a solution

Solving for gradient f and g was no prob:

[tex]\nabla f = <-2x,1>[/tex]

[tex]\nabla g = <2x,2y>[/tex]

And this helps me setup a system of equations:

[tex]1) -2x = 2\lambda x[/tex]

[tex]2) 1 = 2\lambda y[/tex]

[tex]3) x^2+y^2 = 1[/tex]

Now at this point my professor "intuited" that with equation 1, to satisfy the equation, either x=0 or lambda = -1. I agree with it but I don't remember having to "intuit" things in a system of equations, I always remember it being a procedural solution. He takes these two conditions and runs with it and ends up finding the answers...

Now I'm sitting here trying to do it old-school procedurally, so I think equation 2 looks good:

[tex]2) y=\frac{1}{2\lambda}[/tex]

plug into 3

[tex]3) x^2 + (\frac{1}{2\lambda})^2 = 1[/tex]

solve for x:

[tex]x = +/-\sqrt{1-\frac{1}{4(\lambda)^2}}[/tex]

putting the positive version in equation 1:

[tex]1) -2(x) = 2 \lambda x[/tex]

[tex]1) -2 = 2 \lambda[/tex]

[tex]1) \lambda = -1[/tex]

I get the same answer with the positive version of x

[tex]1) \lambda = -1[/tex]

I go back to equation 2 to solve for Y:

[tex] 2) 1 = -2 y[/tex]

[tex] 2) -\frac{1}{2}=y[/tex]

Plugging lambda into my x = sqrt equation for x:

[tex]x = +/- \sqrt{\frac{3}{4}}[/tex]

So now I feel all warm and fuzzy, because I have my extrema locations:

[tex](-\sqrt{\frac{3}{4}}, -\frac{1}{2}); (\sqrt{\frac{3}{4}}, -\frac{1}{2})[/tex]

But looking back at my notes...the x=0 intuition that my professor deduced ended up being another extrema, which ended up leading to the ACTUAL maximum.

Did I screw something up? Is that "intuition" thing actually what I need to do to solve these things? Are there certain conditions that always require this kind of thinking? Is there an even better way to truly find all the solutions?

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# Homework Help: Lagrange Multipliers/System of Equations?

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