高等数学 A2 周维祺
求z=f(x,y)z=f(x,y)z=f(x,y)在约束条件φ(x,y)=0\varphi(x,y)=0φ(x,y)=0下的极值
例:求周长一定的矩形面积的最大值
设矩形边长x,yx,yx,y,周长固定为2a2a2a
f(x,y)=xy,φ(x,y)=x+y−af(x,y)=xy, \varphi(x,y)=x+y-af(x,y)=xy,φ(x,y)=x+y−a
φ(x,y)=0\varphi(x,y)=0φ(x,y)=0的切线方向与f(x,y)f(x,y)f(x,y)等值线的切线方向一致
法线方向即梯度
设(x0,y0)(x_0,y_0)(x0,y0)是z=f(x,y)z=f(x,y)z=f(x,y)在条件φ(x,y)=0\varphi(x,y)=0φ(x,y)=0下的极值点
设梯度存在且不为000,则求解以下方程得x0,y0x_0,y_0x0,y0: {∇f(x0,y0)=λ∇φ(x0,y0)φ(x0,y0)=0\begin{cases}\nabla f(x_0,y_0)=\lambda\nabla \varphi(x_0,y_0) \\\varphi(x_0,y_0)=0 \end{cases}{∇f(x0,y0)=λ∇φ(x0,y0)φ(x0,y0)=0
∇f(x0,y0)=λ∇φ(x0,y0)\nabla f(x_0,y_0)=\lambda\nabla \varphi(x_0,y_0)∇f(x0,y0)=λ∇φ(x0,y0) 即 {fx(x0,y0)=λφx(x0,y0)fy(x0,y0)=λφy(x0,y0)\begin{cases}f_x(x_0,y_0)=\lambda\varphi_x(x_0,y_0) \\ f_y(x_0,y_0)=\lambda\varphi_y(x_0,y_0) \end{cases}{fx(x0,y0)=λφx(x0,y0)fy(x0,y0)=λφy(x0,y0)
设矩形边长x,yx,yx,y,周长固定为2a2a2a,求其面积的最大值
{(y,x)=λ(1,1)x+y−a=0⇒x=y=λ=a/2\begin{cases}(y,x)=\lambda(1,1) \\ x+y-a=0 \end{cases}\quad \Rightarrow\quad x=y=\lambda=a/2{(y,x)=λ(1,1)x+y−a=0⇒x=y=λ=a/2
验证该点是最大值点
在条件x2+y2=1x^2+y^2=1x2+y2=1下分别求:
f(x,y)=x+yf(x,y)=x+yf(x,y)=x+y 的所有极值点
g(x,y)=(x+y)2g(x,y)=(x+y)^2g(x,y)=(x+y)2的所有极值点
在条件x2+y2=3x^2+y^2=3x2+y2=3下求f(x,y)=xyf(x,y)=xyf(x,y)=xy的极值和最值
构造函数L(x,y,λ)=f(x,y)±λφ(x,y)L(x,y,\lambda)=f(x,y)\pm\lambda\varphi(x,y)L(x,y,λ)=f(x,y)±λφ(x,y)
极值点满足:Lx(x,y,λ)=Ly(x,y,λ)=Lλ(x,y,λ)=0L_x(x,y,\lambda)=L_y(x,y,\lambda)=L_{\lambda}(x,y,\lambda)=0Lx(x,y,λ)=Ly(x,y,λ)=Lλ(x,y,λ)=0