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？
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
\[
\left(
\begin{array}{ccccc}
-1 & -1 & 0 & 0 & \frac{3}{2} \\
1 & -\frac{1}{3} & \frac{2}{3} & \frac{2}{3} & -\frac{3}{2} \\
-1 & -\frac{1}{3} & -\frac{1}{3}-\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}-\frac{1}{3} & \frac{3}{2} \\
-1 & -\frac{1}{3} & \frac{1}{\sqrt{3}}-\frac{1}{3} & -\frac{1}{3}-\frac{1}{\sqrt{3}} & \frac{3}{2} \\
-1 & \frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} & \frac{3}{2} \\
1 & \frac{1}{3} & \frac{1}{3}-\frac{1}{\sqrt{3}} & \frac{1}{3}+\frac{1}{\sqrt{3}} & -\frac{3}{2} \\
1 & \frac{1}{3} & \frac{1}{3}+\frac{1}{\sqrt{3}} & \frac{1}{3}-\frac{1}{\sqrt{3}} & -\frac{3}{2} \\
1 & 1 & 0 & 0 & -\frac{3}{2} \\
0 & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 \\
-1 & -\frac{1}{\sqrt{3}} & \frac{1}{6} \left(-3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{1}{6} \left(3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{3}{2} \\
-1 & -\frac{1}{\sqrt{3}} & \frac{1}{6} \left(3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{1}{6} \left(-3^{3/4} \sqrt{2}+\sqrt{3}-3\right) & \frac{3}{2} \\
0 & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 \\
1 & \frac{1}{\sqrt{3}} & \frac{1}{6} \left(-3^{3/4} \sqrt{2}-\sqrt{3}+3\right) & \sqrt{\frac{1}{3} \left(\sqrt{2 \sqrt{3}-3}+1\right)} & -\frac{3}{2} \\
1 & \frac{1}{\sqrt{3}} & \frac{1}{6} \left(3^{3/4} \sqrt{2}-\sqrt{3}+3\right) & \frac{1}{6} \left(-3^{3/4} \sqrt{2}-\sqrt{3}+3\right) & -\frac{3}{2} \\
\end{array}
\right)
\] 穷举法万岁!数值解万岁!
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 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？
Clear["Global`*"];(*Clear all variables*)
f=x^3+y^3+z^3-3*x*y*z+a*(x^2+y^2+z^2-1)
fx=D
fy=D
fz=D
fa=D
out=NSolve[{fx==0,fy==0,fz==0,fa==0},{x,y,z,a}]
f/.out
mathematica对这个问题无能为力 mathematica 发表于 2019-1-9 15:28
mathematica对这个问题无能为力
Clear["Global`*"];(*Clear all variables*)
f1=ContourPlot3D
f2=ContourPlot3D
Show
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？
{x^2+y^2+z^2,x^3+y^3+z^3-3 x y z}//SymmetricReduction[#,{x,y,z},{a,b,c}][]&/@#&//#[]/.Solve[#[]==1][]&//Simplify//(Print[{#,#/.{a->x+y+z}}];Plot[#,{a,-3,3}])&
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 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 ...
Clear["Global`*"];(*Clear all variables*)
Solve[{x^3+y^3+z^3-3*x*y*z==1,x^2+y^2+z^2==1},{x,y}]//FullSimplify//MatrixForm
看结果像是圆的方程
\[\left\{\left\{x\to \frac{1}{2} \left(-z-\sqrt{-z (3 z+4)-2}-2\right),y\to \frac{1}{2} \left(-z+\sqrt{-z (3 z+4)-2}-2\right)\right\},\left\{x\to \frac{1}{2} \left(-z+\sqrt{-z (3 z+4)-2}-2\right),y\to \frac{1}{2} \left(-z-\sqrt{-z (3 z+4)-2}-2\right)\right\},\left\{x\to \frac{1}{2} \left(-z-\sqrt{z (2-3 z)+1}+1\right),y\to \frac{1}{2} \left(-z+\sqrt{z (2-3 z)+1}+1\right)\right\},\left\{x\to \frac{1}{2} \left(-z+\sqrt{z (2-3 z)+1}+1\right),y\to \frac{1}{2} \left(-z-\sqrt{z (2-3 z)+1}+1\right)\right\}\right\}\] 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？ Clear["Global`*"];(*Clear all variables*)
out=FullSimplify@Solve[{x^3+y^3+z^3-3*x*y*z==1,x^2+y^2+z^2==1},{x,y}]
f1=ParametricPlot[{x,y}/.out[],{z,-1/3,1}]
f2=ParametricPlot[{x,y}/.out[],{z,-1/3,1}]
Show
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？ 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{\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{\sqrt{f^2-1}-f}+\frac{1}{\sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{1}{\sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{1}{\sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{1}{\sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}+\frac{\left(\sqrt{3}+i\right) i}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt{\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{\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{\sqrt{f^2-1}-f}-\frac{1+i \sqrt{3}}{2 \sqrt{\sqrt{f^2-1}-f}}-x\right)\right\}\)
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