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

Find the extrema of the function subject to the given constraint.

f (x, y) = x

x^{2}+ 2y^{2}= 3

2. Relevant equations

det = f_{xx}* f_{yy}- (f_{xy})^{2}

If det > 0,

f_{xx}< 0 [tex]\Rightarrow[/tex] MAXIMUM

f_{xx}> 0 [tex]\Rightarrow[/tex] MINIMUM

If det < 0,

[tex]\Rightarrow[/tex] SADDLE POINT

If det = 0m

Inconclusive

3. The attempt at a solution

I appear to be struggling with this maxima/minima stuff. Previous questions werejustcorrect. I had to look at the book's answers to figure a way there. I seem to be missing something...but here is my attempt:

Since x^{2}+ 2y^{2}= 3,

x = [tex]\pm[/tex][tex]\sqrt{3 - 2y^{2}}[/tex]

Therefore,

f(x, y) = [tex]\pm[/tex][tex]\sqrt{3 - 2y^{2}}[/tex]

At critical point,

[tex]\nabla[/tex] f = (0, - [tex]\frac{2y}{\sqrt{3 - 2y^{2}}}[/tex] )

= (0, 0)

Therefore, equating gives

0 = 0

&

- [tex]\frac{2y}{\sqrt{3 - 2y^{2}}}[/tex] = 0

So extrema points at([tex]\sqrt{3}[/tex], 0)&(-[tex]\sqrt{3}[/tex], 0)since y = 0.

To find max/min/saddle, find det.

f_{xx}= 0

f_{xy}= 0

det = f_{xx}* f_{yy}- (f_{xy})^{2}

det = 0 * f_{yy}- (0)^{2}

det = 0

Which is inconclusive. The book has ([tex]\sqrt{3}[/tex], 0) as a maximum and (-[tex]\sqrt{3}[/tex], 0) as a minimum.

I don't know what I'm missing out here...

Thanks in advance.

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# Homework Help: Maxima/Minima of extrema function f(x,y)=x

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