Respuesta :
The Lagrangian
[tex]L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)[/tex]
has critical points where the first derivatives vanish:
[tex]L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}[/tex]
[tex]L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}[/tex]
[tex]L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}[/tex]
[tex]L_\lambda=x^4+y^4+z^4-13=0[/tex]
We can't have [tex]x=y=z=0[/tex], since that contradicts the last condition.
(0 critical points)
If two of them are zero, then the remaining variable has two possible values of [tex]\pm\sqrt[4]{13}[/tex]. For example, if [tex]y=z=0[/tex], then [tex]x^4=13\implies x=\pm\sqrt[4]{13}[/tex].
(6 critical points; 2 for each non-zero variable)
If only one of them is zero, then the squares of the remaining variables are equal and we would find [tex]\lambda=-\frac1{\sqrt{26}}[/tex] (taking the negative root because [tex]x^2,y^2,z^2[/tex] must be non-negative), and we can immediately find the critical points from there. For example, if [tex]z=0[/tex], then [tex]x^4+y^4=13[/tex]. If both [tex]x,y[/tex] are non-zero, then [tex]x^2=y^2=-\frac1{2\lambda}[/tex], and
[tex]xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}[/tex]
[tex]\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}[/tex]
and for either choice of [tex]x[/tex], we can independently choose from [tex]y=\pm\sqrt[4]{\frac{13}2}[/tex].
(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)
If none of the variables are zero, then [tex]x^2=y^2=z^2=-\frac1{2\lambda}[/tex]. We have
[tex]xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}[/tex]
[tex]\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}[/tex]
and similary [tex]y,z[/tex] have the same solutions whose signs can be picked independently of one another.
(8 critical points)
Now evaluate [tex]f[/tex] at each critical point; you should end up with a maximum value of [tex]\sqrt{39}[/tex] and a minimum value of [tex]\sqrt{13}[/tex] (both occurring at various critical points).
Here's a comprehensive list of all the critical points we found:
[tex](\sqrt[4]{13},0,0)[/tex]
[tex](-\sqrt[4]{13},0,0)[/tex]
[tex](0,\sqrt[4]{13},0)[/tex]
[tex](0,-\sqrt[4]{13},0)[/tex]
[tex](0,0,\sqrt[4]{13})[/tex]
[tex](0,0,-\sqrt[4]{13})[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)[/tex]
[tex]\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)[/tex]
