求助一道最值问题

.Sun

求助，想了很长时间也没什么头绪

northwolves

利用60度角对应的△ABC,△ADE,△ADF，三个余弦定理理论上可以解出

mathematica

1. Clear["Global`*"];
   
2. (*计算余弦值子函数,利用三边计算余弦值*)
   
3. cs[a_,b_,c_]:=(a^2+b^2-c^2)/(2*a*b)
   
4. FullSimplify@Maximize[
   
5.     {DF,
   
6.         (*假设BD=EC=FC=x,则BC=4*x*)
   
7.         1/2==cs[AD,AE+2*x,DF]&&
   
8.         1/2==cs[AD,AE,2]&&
   
9.         1/2==cs[AD+x,AE+x,4*x]&&
   
10.         AD>0&&
    
11.         AE>0&&
    
12.         DF>0&&
    
13.         x>0
    
14.     },{AD,AE,DF,x}
    
15. ]
    
16. N[%]
    

复制代码

三次利用余弦定理
求解结果如下：
\[\left\{\frac{2}{15} \left(\sqrt{259}+8\right),\left\{\text{AD}\to \sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}},\text{AE}\to \sqrt{\frac{1292}{307}+\frac{4048}{307 \sqrt{259}}},\text{DF}\to \frac{2}{15} \left(\sqrt{259}+8\right),x\to \frac{2}{15} \sqrt{17+\frac{76}{\sqrt{259}}}\right\}\right\}\]
数值如下
{3.21246,{AD->0.64243,AE->2.24227,DF->3.21246,x->0.621431}}

mathematica

你缺的不是头绪，你缺的是计算软件！

mathematica

1. Clear["Global`*"];
   
2. (*计算余弦值子函数,角度60*)
   
3. fun[a_,b_,c_]:=a^2+b^2-2*a*b*(1/2)-c^2
   
4. (*三个约数条件*)
   
5. f1=fun[AD,AE,2]
   
6. f2=fun[AD+x,AE+x,4*x]
   
7. f3=fun[AD,AE+2*x,DF]
   
8. (*拉格朗日乘子法目标函数+约数条件*)
   
9. yy=DF+x1*f1+x2*f2+x3*f3
   
10. (*求解七个偏导数*)
    
11. yyAD=D[yy,AD]
    
12. yyAE=D[yy,AE]
    
13. yyDF=D[yy,DF]
    
14. yyx=D[yy,x]
    
15. yyx1=D[yy,x1]
    
16. yyx2=D[yy,x2]
    
17. yyx3=D[yy,x3]
    
18. (*联立求解方程组*)
    
19. out=Solve[
    
20. yyAD==0&&
    
21. yyAE==0&&
    
22. yyDF==0&&
    
23. yyx==0&&
    
24. yyx1==0&&
    
25. yyx2==0&&
    
26. yyx3==0,
    
27. {AD,AE,DF,x,x1,x2,x3}
    
28. ];
    
29. out1=FullSimplify[out]
    
30. (*转化成根式显式表达*)
    
31. Grid@ToRadicals[out1]
    
32. N[%]
    

复制代码

求解结果
\[\begin{array}{ccccccc}
\text{AD}\to -\sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to -2 \sqrt{\frac{323}{307}+\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}+8\right) & x\to \frac{1}{15} (-2) \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}+8\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(8-\sqrt{259}\right) \\
\text{AD}\to -\sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to -2 \sqrt{\frac{323}{307}+\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}+8\right) & x\to \frac{1}{15} (-2) \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(-\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}-8\right) \\
\text{AD}\to \sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to \sqrt{\frac{1292}{307}+\frac{4048}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}+8\right) & x\to \frac{2}{15} \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}+8\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(8-\sqrt{259}\right) \\
\text{AD}\to \sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to \sqrt{\frac{1292}{307}+\frac{4048}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}+8\right) & x\to \frac{2}{15} \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(-\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}-8\right) \\
\text{AD}\to -2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to 2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}-8\right) & x\to \frac{1}{15} (-2) \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(-\sqrt{259}-8\right) \\
\text{AD}\to -2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to 2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}-8\right) & x\to \frac{1}{15} (-2) \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(8-\sqrt{259}\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}+8\right) \\
\text{AD}\to 2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to -2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}-8\right) & x\to \frac{2}{15} \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(-\sqrt{259}-8\right) \\
\text{AD}\to 2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to -2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}-8\right) & x\to \frac{2}{15} \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(8-\sqrt{259}\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}+8\right) \\
\end{array}\]

\[\begin{array}{ccccccc}
\text{AD}\to -0.64243 & \text{AE}\to -2.24227 & \text{DF}\to -3.21246 & x\to -0.621431 & \text{x1}\to 0.401558 & \text{x2}\to -0.125 & \text{x3}\to -0.155644 \\
\text{AD}\to -0.64243 & \text{AE}\to -2.24227 & \text{DF}\to 3.21246 & x\to -0.621431 & \text{x1}\to -0.401558 & \text{x2}\to 0.125 & \text{x3}\to 0.155644 \\
\text{AD}\to 0.64243 & \text{AE}\to 2.24227 & \text{DF}\to -3.21246 & x\to 0.621431 & \text{x1}\to 0.401558 & \text{x2}\to -0.125 & \text{x3}\to -0.155644 \\
\text{AD}\to 0.64243 & \text{AE}\to 2.24227 & \text{DF}\to 3.21246 & x\to 0.621431 & \text{x1}\to -0.401558 & \text{x2}\to 0.125 & \text{x3}\to 0.155644 \\
\text{AD}\to -0.287051 & \text{AE}\to 1.84096 & \text{DF}\to -1.07913 & x\to -0.467192 & \text{x1}\to 0.134891 & \text{x2}\to 0.125 & \text{x3}\to -0.463336 \\
\text{AD}\to -0.287051 & \text{AE}\to 1.84096 & \text{DF}\to 1.07913 & x\to -0.467192 & \text{x1}\to -0.134891 & \text{x2}\to -0.125 & \text{x3}\to 0.463336 \\
\text{AD}\to 0.287051 & \text{AE}\to -1.84096 & \text{DF}\to -1.07913 & x\to 0.467192 & \text{x1}\to 0.134891 & \text{x2}\to 0.125 & \text{x3}\to -0.463336 \\
\text{AD}\to 0.287051 & \text{AE}\to -1.84096 & \text{DF}\to 1.07913 & x\to 0.467192 & \text{x1}\to -0.134891 & \text{x2}\to -0.125 & \text{x3}\to 0.463336 \\
\end{array}\]

.Sun

谢谢大家，现在找到答案了，属于阿氏圆最值问题

mathematica

.Sun 发表于 2020-3-15 17:26
谢谢大家，现在找到答案了，属于阿氏圆最值问题

奇技淫巧！
代数、微积分的办法，多么自然而然的思路呀！

王守恩

谢谢 mathematica！依样画葫芦：FullSimplify@Maximize

换已知条件 DE=2 为 DE=sin60°，  记 DA=sin(x)，DB=sin(y)

可得
\((\sin(x)+\sin(y))^2+(\sin(120^\circ-x)+\sin(y))^2-2(\sin(x)+\sin(y))(\sin(120^\circ-x)+\sin(y))\cos(60^\circ)=(4\sin(y))^2\)
求\(\ \ \sqrt{(\sin(60^\circ))^2+(2\sin(y))^2+2\sin(60^\circ)(2\sin(y))\cos(x)}\ \ \)的最大值

化简
\(30\cos(2y)+2\sqrt{3}\cos(x)\sin(y)+6\sin(x)\sin(y)=27\)
求\(\ \ \sqrt{\frac{3}{4}+2\sqrt{3}\cos(x)\sin(y)+(2\sin(y))^2}\ \ \)的最大值

mathematica

王守恩 发表于 2020-3-17 08:38
谢谢 mathematica！依样画葫芦：FullSimplify@Maximize

换已知条件 DE=2 为 DE=sin60°，  记 DA=sin(x) ...

对你的不感兴趣！
我只对有推理过程的感兴趣，
其余乱猜测胡乱说的，不感兴趣