MHB Travis Henderson's Question: Optimizing f(x,y,z) with Constraint

AI Thread Summary
The objective function f(x,y,z)=x^4+y^4+z^4 is analyzed under the constraint x^2+y^2+z^2=1 using Lagrange multipliers. Critical points are identified, resulting in 12 points where one variable is zero and the other two are equal, yielding a function value of 1/2. Additionally, there are 6 points where two variables are zero and one is ±1, giving a value of 1. Finally, 8 points arise from equal variables, leading to a minimum value of 1/3. The maximum value is confirmed to be 1, while the minimum is 1/3.
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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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