在x^2+y^2+z^2=1的条件下，如何求得f=x^3+y^3+z^3-3*x*y*z的最值分布情况？

mathematica

在 \(x^2+y^2+z^2=1\) 的条件下，如何求得 \(f=x^3+y^3+z^3-3xyz\) 的最值分布情况？
哪些情况下可以求得最大值，最大值是哪些点？
哪些情况下可以求得最小值，最小值是哪些点？

注意：是所有的最大值点，也就是哪些情况下取最大值，哪些情况下取最小值

\[
\left(
\begin{array}{ccccc}
-1 & -1 & 0 & 0 & \frac{3}{2} \\
1 & -\frac{1}{3} & \frac{2}{3} & \frac{2}{3} & -\frac{3}{2} \\
-1 & -\frac{1}{3} & -\frac{1}{3}-\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}-\frac{1}{3} & \frac{3}{2} \\
-1 & -\frac{1}{3} & \frac{1}{\sqrt{3}}-\frac{1}{3} & -\frac{1}{3}-\frac{1}{\sqrt{3}} & \frac{3}{2} \\
-1 & \frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} & \frac{3}{2} \\
1 & \frac{1}{3} & \frac{1}{3}-\frac{1}{\sqrt{3}} & \frac{1}{3}+\frac{1}{\sqrt{3}} & -\frac{3}{2} \\
1 & \frac{1}{3} & \frac{1}{3}+\frac{1}{\sqrt{3}} & \frac{1}{3}-\frac{1}{\sqrt{3}} & -\frac{3}{2} \\
1 & 1 & 0 & 0 & -\frac{3}{2} \\
0 & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 \\
-1 & -\frac{1}{\sqrt{3}} & \frac{1}{6} \left(-3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{1}{6} \left(3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{3}{2} \\
-1 & -\frac{1}{\sqrt{3}} & \frac{1}{6} \left(3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{1}{6} \left(-3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{3}{2} \\
0 & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 \\
1 & \frac{1}{\sqrt{3}} & \frac{1}{6} \left(-3^{3/4} \sqrt{2}-\sqrt{3}+3\right) & \sqrt{\frac{1}{3} \left(\sqrt{2 \sqrt{3}-3}+1\right)} & -\frac{3}{2} \\
1 & \frac{1}{\sqrt{3}} & \frac{1}{6} \left(3^{3/4} \sqrt{2}-\sqrt{3}+3\right) & \frac{1}{6} \left(-3^{3/4} \sqrt{2}-\sqrt{3}+3\right) & -\frac{3}{2} \\
\end{array}
\right)
\]

zeroieme

穷举法万岁!数值解万岁!

mathematica

zeroieme 发表于 2019-1-9 14:59
穷举法万岁!数值解万岁!

$x^3+y^3+z^3-3 x y z=(x+y+z)^3-3 (x+y+z) (x y+x z+y z)$
看你怎么秒破？

mathematica

mathematica 发表于 2019-1-9 15:19
$x^3+y^3+z^3-3 x y z=(x+y+z)^3-3 (x+y+z) (x y+x z+y z)$
看你怎么秒破？

1. Clear["Global`*"];(*Clear all variables*)
   
2. f=x^3+y^3+z^3-3*x*y*z+a*(x^2+y^2+z^2-1)
   
3. fx=D[f,x]
   
4. fy=D[f,y]
   
5. fz=D[f,z]
   
6. fa=D[f,a]
   
7. out=NSolve[{fx==0,fy==0,fz==0,fa==0},{x,y,z,a}]
   
8. f/.out
   

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mathematica对这个问题无能为力

mathematica

mathematica 发表于 2019-1-9 15:28
mathematica对这个问题无能为力

1. Clear["Global`*"];(*Clear all variables*)
   
2. f1=ContourPlot3D[x^3+y^3+z^3-3*x*y*z==1,{x,-1,1},{y,-1,1},{z,-1,1}]
   
3. f2=ContourPlot3D[x^2+y^2+z^2==1,{x,-1,1},{y,-1,1},{z,-1,1},Mesh->Mone]
   
4. Show[f1,f2]
   

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此处是把两个都画出来，谁能把共同部分画出来？
可以使用软件求出最大值是1
谁能把共同部分画出来？

zeroieme

1. {x^2+y^2+z^2,x^3+y^3+z^3-3 x y z}//SymmetricReduction[#,{x,y,z},{a,b,c}][[1]]&/@#&//#[[2]]/.Solve[#[[1]]==1][[1]]&//Simplify//(Print[{#,#/.{a->x+y+z}}];Plot[#,{a,-3,3}])&

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把Mathematica当VB写的mathematica当然对这个问题无能为力，因为不懂三次多项式根本就没有最大值最小值。

mathematica

mathematica 发表于 2019-1-9 15:41
此处是把两个都画出来，谁能把共同部分画出来？
可以使用软件求出最大值是1
谁能把共同部分画出来 ...

1. Clear["Global`*"];(*Clear all variables*)
   
2. Solve[{x^3+y^3+z^3-3*x*y*z==1,x^2+y^2+z^2==1},{x,y}]//FullSimplify//MatrixForm
   

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看结果像是圆的方程
\[\left\{\left\{x\to \frac{1}{2} \left(-z-\sqrt{-z (3 z+4)-2}-2\right),y\to \frac{1}{2} \left(-z+\sqrt{-z (3 z+4)-2}-2\right)\right\},\left\{x\to \frac{1}{2} \left(-z+\sqrt{-z (3 z+4)-2}-2\right),y\to \frac{1}{2} \left(-z-\sqrt{-z (3 z+4)-2}-2\right)\right\},\left\{x\to \frac{1}{2} \left(-z-\sqrt{z (2-3 z)+1}+1\right),y\to \frac{1}{2} \left(-z+\sqrt{z (2-3 z)+1}+1\right)\right\},\left\{x\to \frac{1}{2} \left(-z+\sqrt{z (2-3 z)+1}+1\right),y\to \frac{1}{2} \left(-z-\sqrt{z (2-3 z)+1}+1\right)\right\}\right\}\]

mathematica

zeroieme 发表于 2019-1-9 15:47
把Mathematica当VB写的mathematica当然对这个问题无能为力，因为不懂三次多项式根本就没有最大值最小值。

谁告诉你这个问题没最大值最小值的？

mathematica

1. Clear["Global`*"];(*Clear all variables*)
   
2. out=FullSimplify@Solve[{x^3+y^3+z^3-3*x*y*z==1,x^2+y^2+z^2==1},{x,y}]
   
3. f1=ParametricPlot[{x,y}/.out[[3]],{z,-1/3,1}]
   
4. f2=ParametricPlot[{x,y}/.out[[4]],{z,-1/3,1}]
   
5. Show[f1,f2,PlotRange->{{-1,1},{-1,1}}]
   

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由这个绘图结果，可以知道在xy平面上的投影是一个椭圆，
那么如何得到这个椭圆的方程呢？

zeroieme

以下参数解已经满足条件\(x^2+y^2+z^2=1\)，f可以取任意大！
\(\left\{y\to \frac{1}{2} \left(-\sqrt{\frac{2 \left(\left(\sqrt{f^2-1}-f\right)^{2/3}+1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\sqrt{f^2-1}-f\right)^{4/3}+1}{\left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2}+\sqrt[3]{\sqrt{f^2-1}-f}+\frac{1}{\sqrt[3]{\sqrt{f^2-1}-f}}-x\right),z\to \frac{1}{2} \left(\sqrt{\frac{2 \left(\left(\sqrt{f^2-1}-f\right)^{2/3}+1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\sqrt{f^2-1}-f\right)^{4/3}+1}{\left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2}+\sqrt[3]{\sqrt{f^2-1}-f}+\frac{1}{\sqrt[3]{\sqrt{f^2-1}-f}}-x\right)\right\},\left\{y\to \frac{1}{2} \left(\sqrt{\frac{2 \left(\left(\sqrt{f^2-1}-f\right)^{2/3}+1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\sqrt{f^2-1}-f\right)^{4/3}+1}{\left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2}+\sqrt[3]{\sqrt{f^2-1}-f}+\frac{1}{\sqrt[3]{\sqrt{f^2-1}-f}}-x\right),z\to \frac{1}{2} \left(-\sqrt{\frac{2 \left(\left(\sqrt{f^2-1}-f\right)^{2/3}+1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\sqrt{f^2-1}-f\right)^{4/3}+1}{\left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2}+\sqrt[3]{\sqrt{f^2-1}-f}+\frac{1}{\sqrt[3]{\sqrt{f^2-1}-f}}-x\right)\right\},\left\{y\to \frac{1}{2} \left(-\sqrt{\frac{\left(-\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}-\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right),z\to \frac{1}{2} \left(\sqrt{\frac{\left(-\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}-\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right)\right\},\left\{y\to \frac{1}{2} \left(\sqrt{\frac{\left(-\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}-\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right),z\to \frac{1}{2} \left(-\sqrt{\frac{\left(-\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1+i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}-\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right)\right\},\left\{y\to \frac{1}{2} \left(-\sqrt{\frac{\left(\left(\sqrt{3}+i\right) i \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1-i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}+\frac{1}{2} \left(\sqrt{3}+i\right) i \sqrt[3]{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right),z\to \frac{1}{2} \left(\sqrt{\frac{\left(\left(\sqrt{3}+i\right) i \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1-i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}+\frac{1}{2} \left(\sqrt{3}+i\right) i \sqrt[3]{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right)\right\},\left\{y\to \frac{1}{2} \left(\sqrt{\frac{\left(\left(\sqrt{3}+i\right) i \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1-i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}+\frac{1}{2} \left(\sqrt{3}+i\right) i \sqrt[3]{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right),z\to \frac{1}{2} \left(-\sqrt{\frac{\left(\left(\sqrt{3}+i\right) i \left(\sqrt{f^2-1}-f\right)^{2/3}-i \sqrt{3}-1\right) x}{\sqrt[3]{\sqrt{f^2-1}-f}}-\frac{\left(\left(1-i \sqrt{3}\right) \left(\sqrt{f^2-1}-f\right)^{2/3}+\sqrt{3} i+1\right)^2}{4 \left(\sqrt{f^2-1}-f\right)^{2/3}}-3 x^2+2}+\frac{1}{2} \left(\sqrt{3}+i\right) i \sqrt[3]{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt[3]{\sqrt{f^2-1}-f}}-x\right)\right\}\)