下的条件极值,用Lagrange乘数法求解
The stagnation point is outside the circle, and there is no stagnation point inside the circle, so the extreme value is not taken. Next consider the highest value on
. This is the conditional extreme value under the constraint bar, which is solved by the Lagrange multiplier method.
In[3]:= Clear[x,y,F,t];F[x_,y_,t_]:=f[x,y] t(x^2 y^2-25);
s=Solve[{D[f[x,y,t]==0,D[f[x,y,t]==0,y],D[F[x,y,t]= =0,t]],{x,y,t}}
Out[3]={{t->-3,x->-3,y->4},{t->1,x->3,y->-4}}
In[4]:= F[x,y]/.s[[1]]
Out[4]=25
In[5]:= F[x,y]/.s[[2]]
Out[5]=-75
今天的分享就到这里啦
That's it for today's sharing
散会
Let's adjourn
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翻译：Google翻译
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