Clearly, f(x,y)=x3y{\displaystyle f(x,y)=x^{3}y} and g(x,y)=3x2+y2=6. {\displaystyle g(x,y)=3x^{2}+y^{2}=6. }
∇f+λ∇g=0{\displaystyle \nabla f+\lambda \nabla g=0} {3x2y+λ6x=0x3+λ2y=0{\displaystyle {\begin{cases}3x^{2}y+\lambda 6x&=0\x^{3}+\lambda 2y&=0\end{cases}}}
{3x2y2+λ6xy=03x4+λ6xy=0{\displaystyle {\begin{cases}3x^{2}y^{2}+\lambda 6xy&=0\3x^{4}+\lambda 6xy&=0\end{cases}}} 3x2y2−3x4=0x2(y2−x2)=0{\displaystyle {\begin{aligned}3x^{2}y^{2}-3x^{4}&=0\x^{2}(y^{2}-x^{2})&=0\end{aligned}}}
y=x{\displaystyle y=x}
g=3x2+x2=6x=y=±32{\displaystyle {\begin{aligned}g&=3x^{2}+x^{2}=6\x&=y=\pm {\sqrt {\frac {3}{2}}}\end{aligned}}}
x3y=94{\displaystyle x^{3}y={\frac {9}{4}}}
L(x,y,z;λ)=(x2+y2+z2)+λ(x2yz−1){\displaystyle {\mathcal {L}}(x,y,z;\lambda )=(x^{2}+y^{2}+z^{2})+\lambda (x^{2}yz-1)}
{∂L∂x=2x+λ2xyz=0∂L∂y=2y+λx2z=0∂L∂z=2z+λx2y=0{\displaystyle {\begin{cases}{\frac {\partial {\mathcal {L}}}{\partial x}}=2x+\lambda 2xyz&=0\{\frac {\partial {\mathcal {L}}}{\partial y}}=2y+\lambda x^{2}z&=0\{\frac {\partial {\mathcal {L}}}{\partial z}}=2z+\lambda x^{2}y&=0\end{cases}}}
{2x2+λ2x2yz=04y2+λ2x2yz=04z2+λ2x2yz=0{\displaystyle {\begin{cases}2x^{2}+\lambda 2x^{2}yz&=0\4y^{2}+\lambda 2x^{2}yz&=0\4z^{2}+\lambda 2x^{2}yz&=0\end{cases}}}
x2=2y2=2z2{\displaystyle x^{2}=2y^{2}=2z^{2}} The equation above gives us all the information we need to optimize the distance now.
(2y2)(y)(y)=1y=2−1/4{\displaystyle {\begin{aligned}(2y^{2})(y)(y)&=1\y&=2^{-1/4}\end{aligned}}}
x2+y2+z2=2y2+y2+y2=4y2=2⋅2−1/4=23/4{\displaystyle {\begin{aligned}{\sqrt {x^{2}+y^{2}+z^{2}}}&={\sqrt {2y^{2}+y^{2}+y^{2}}}\&={\sqrt {4y^{2}}}\&=2\cdot 2^{-1/4}\&=2^{3/4}\end{aligned}}}
