| |
题：已知$\,x^2+\frac{1}{\large x}+y^2+\frac{1}{\large y}=\frac{27}{4}$，求$\,{\small P}=\frac{15}{\large x}–\frac{3}{4y}\,$的最小值. srcl<
解：应用Lagrange乘数法,求AFz6
$\qquad L(x,y,\lambda)=\frac{15}{x}-\frac{3}{4y}+\lambda(x^2+\frac{1}{\large x}+y^2+\frac{1}{\large y}-\frac{27}{4})\,$的驻点$\scriptsize(\text{Station Point})$.G\K;!7
$\underset{\,}{\qquad}$令$\,(L_x,L_y,L_{\lambda})\equiv\big({\small-}\frac{15+\lambda}{x^2}{\small +2\lambda x},\frac{3-4\lambda}{4y^2}{\small +2\lambda y},{\small x^2+}\frac{1}{x}+{\small y^2+}\frac{1}{y}-\frac{27}{4}\big)=\mathbf{0}$.!Ag
$\underset{\,}{\qquad}$得$\;(x,y)=\big(\sqrt[3]{\frac{1}{2}+\frac{15}{2\lambda}},\,\sqrt[3]{\frac{1}{2}-\frac{3}{8\lambda}}\big),\;\small\;4y(x^3+1)+4x(y^3+1)=27xy.$aL
$\underset{\,}{\quad}$从 ScientificWorkplace 得到,又由Pari/GP矫正的数值解是#u26
$\underset{\,}{\quad}\lambda\scriptsize =-15.2733775542663202790335668352041186011244897907934778446722906011405589445447574850433\ldots$X
$\underset{\,}{\quad}x_{\lambda}\scriptsize =0.2076184400085877397903182624732172770279907559896122126763520376218531613309925423184489\ldots$^X%S
$\underset{\,}{\quad}y_{\lambda}\scriptsize =0.8064850618904549111593181843820115312225902750151762121200379377395120529744293547350502\ldots$:7))\B
$\underset{\,}{\quad}\frac{15}{x_{\lambda}}-\frac{3}{8y_{\lambda}}\scriptsize =71.78293714607018550281334807129473917486703810592176872136249966161294571285335719\ldots$<L%S
 KN
| | |
|
|
|
顶端 
加到"个人收藏夹"
主题管理：
总固顶 
取消总固顶 
消息
查看
搜索
好友
引用
回复
只看我
2018/05/15 08:34am 
IP: 已设置保密
