初中题，三角形ABC中，BC=8.∠A=60°，如何求AB+AC/2的最大值？

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. Maximize[{AB+1/2*AC,cs[AB,AC,8]==Cos[60*Degree]},{AB,AC}]//FullSimplify
   

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\[\left\{8 \sqrt{\frac{7}{3}},\left\{\text{AB}\to \frac{40}{\sqrt{21}},\text{AC}\to \frac{32}{\sqrt{21}}\right\}\right\}\]

mathematica

mathematica 发表于 2020-6-16 09:08
\[\left\{8 \sqrt{\frac{7}{3}},\left\{\text{AB}\to \frac{40}{\sqrt{21}},\text{AC}\to \frac{32}{\s ...

1. Clear["Global`*"];
   
2. ans=Solve[{x+y/2==k,x^2-x*y+y^2==64},{x,y}]//FullSimplify
   

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求解结果
\[\begin{array}{cc}
x\to \frac{1}{7} \left(5 k-\sqrt{448-3 k^2}\right) & y\to \frac{2}{7} \left(\sqrt{448-3 k^2}+2 k\right) \\
x\to \frac{1}{7} \left(\sqrt{448-3 k^2}+5 k\right) & y\to \frac{1}{7} (-2) \left(\sqrt{448-3 k^2}-2 k\right) \\
\end{array}\]

mathematica

mathematica 发表于 2020-6-16 09:10
求解结果
\[\begin{array}{cc}
x\to \frac{1}{7} \left(5 k-\sqrt{448-3 k^2}\right) & y\to \frac{ ...

1. Clear["Global`*"];
   
2. f=x+y/2+t*(x^2-x*y+y^2-64)
   
3. (*拉格朗日乘子法计算极值点*)
   
4. ans=Solve[D[f,{{x,y,t}}]==0,{x,y,t}]//FullSimplify
   
5. (*带入极值点求解最值*)
   
6. f/.ans//FullSimplify
   

复制代码

\[\left\{\left\{x\to -\frac{40}{\sqrt{21}},y\to -\frac{32}{\sqrt{21}},t\to \frac{\sqrt{\frac{7}{3}}}{16}\right\},\left\{x\to \frac{40}{\sqrt{21}},y\to \frac{32}{\sqrt{21}},t\to -\frac{\sqrt{\frac{7}{3}}}{16}\right\}\right\}\]

\[\left\{-8 \sqrt{\frac{7}{3}},8 \sqrt{\frac{7}{3}}\right\}\]