(*计算余弦值子函数,利用三边计算余弦值*)
cs:=(a^2+b^2-c^2)/(2*a*b)
Maximize[{AB+1/2*AC,cs==Cos},{AB,AC}]//FullSimplify
\[\left\{8 \sqrt{\frac{7}{3}},\left\{\text{AB}\to \frac{40}{\sqrt{21}},\text{AC}\to \frac{32}{\sqrt{21}}\right\}\right\}\] 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 ...
Clear["Global`*"];
ans=Solve[{x+y/2==k,x^2-x*y+y^2==64},{x,y}]//FullSimplify
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 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{ ...
Clear["Global`*"];
f=x+y/2+t*(x^2-x*y+y^2-64)
(*拉格朗日乘子法计算极值点*)
ans=Solve==0,{x,y,t}]//FullSimplify
(*带入极值点求解最值*)
f/.ans//FullSimplify
\[\left\{\left\{x\to -\frac{40}{\sqrt{21}},y\to -\frac{32}{\sqrt{21}},t\to \frac{\sqrt{\frac{7}{3}}}{16}\right\},\left\{x\to \frac{40}{\sqrt{21}},y\to \frac{32}{\sqrt{21}},t\to -\frac{\sqrt{\frac{7}{3}}}{16}\right\}\right\}\]
\[\left\{-8 \sqrt{\frac{7}{3}},8 \sqrt{\frac{7}{3}}\right\}\]
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