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Travis Henderson's question at Yahoo! Answers regarding optimization with constraint

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MarkFL

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Feb 24, 2012
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MarkFL

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Feb 24, 2012
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Hello Travis,

We are given the objective function:

$f(x,y,z)=x^4+y^4+z^4$

subject to the constraint:

$g(x,y,z)=x^2+y^2+z^2-1=0$

Using Lagrange multipliers, we obtain the system:

$4x^3=\lambda(2x)$

$4y^3=\lambda(2y)$

$4z^3=\lambda(2z)$

We see that 12 critical points arise when one of the variables is zero, and the other two are not zero. We see that the other two have to be equal, and their value is found from the constraint:

$y^2+x^2=1$

$x=y=\pm\frac{1}{\sqrt{2}}$

The 12 critical points come from the permutations of:

$\displaystyle \left(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}} \right),\,\left(\pm\frac{1}{\sqrt{2}},0,\pm\frac{1}{\sqrt{2}} \right),\,\left(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}},0 \right)$

The objective function's value is the same at each of the 12 points and is given by:

$f_1=\dfrac{1}{2}$

We also see that there are 6 critical values that arise from two of the varaibles being zero, and the other one being $\pm1$. They are:

$(0,0,\pm1),\,(0,\pm1,0),\,(\pm1,0,0)$

The objective function's value is the same at each of the 12 points and is given by:

$f_2=1$

Lastly the other 8 critical values comes from:

$x=y=z$

and substituting into the constraint, we find:

$x=y=z=\pm\dfrac{1}{\sqrt{3}}$

and so we have the 8 permutations of:

$f_3=f\left(\pm\dfrac{1}{\sqrt{3}},\pm\dfrac{1}{ \sqrt{3}},\pm\dfrac{1}{\sqrt{3}} \right)=\dfrac{1}{3}$

Hence we find:

$f_{\text{min}}=\dfrac{1}{3}$

$f_{\text{max}}=1$
 
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