6.7 2変数関数の極値

1.

2変数関数のTaylorの定理

$$f(x_{0}+h,y_{0}+k) = f(x_{0},y_{0}) + (h\frac{\partial}{\partial x} + k\frac{\partial}... ...\partial}{\partial x} + k\frac{\partial}{\partial y})^2 f(x_{0},y_{0}) + \cdots$$

(a)Taylorの定理で $x_{0} = 1, y_{0}=1, 1+h = x, 1+k = y$とおくと

$$f(x,y) = f(1,1) + ((x-1)\frac{\partial}{\partial x} + (y-1)\frac{... ...rac{\partial}{\partial x} + (y-1)\frac{\partial}{\partial y})^2 f(1,1) + \cdots$$

$$f_{x} = 2xy, f_{y} = x^2$$

$$f_{xx} = 2y, f_{xy} = 2x, f_{yy} = 0$$

よって

$$f(1,1) = 1, f_{x}(1,1) = 2, f_{y}(1,1) = 1, f_{xx}(1,1) = 2, f_{xy}(1,1) = 2, f_{yy}(1,1) = 0$$

したがって，

$$f(x,y) = 1 + 2(x-1) + y-1 + \frac{1}{2}\left(2(x-1)^{2} + 4(x-1)(y-1)\right)$$

$$= 1 + 2(x-1) + y-1 + (x-1)^2 + 2(x-1)(y-1)$$

(b)

$$f_{x} = -y\sin{xy}, f_{y} = -x\sin{xy}$$

$$f_{xx} = -y^{2}\cos{xy}, f_{xy} = -\sin{xy} -xy\cos{xy}, f_{yy} = -x^{2}\cos{xy}$$

よって

$$\begin{array}{ll} f(1,\frac{\pi}{2}) = \cos{\frac{\pi}{2}} = ... ...,\frac{\pi}{2}) = -1 & f_{yy}(1,\frac{\pi}{2}) = -1 \end{array}$$

したがって，

$$f(x,y) = 0 - \frac{\pi}{2}(x-1) - (y - \frac{\pi}{2})$$

$$+ \frac{1}{2}\left((x-1)^{2}(-\frac{\pi^{2}}{4}) + 2(x-1)(y - \frac{\pi}{2})(-1) + (-1)(y - \frac{\pi}{2})^{2}\right)$$

(c)

$$f_{x} = \frac{-1}{1 - x+y}, f_{y} = \frac{1}{1 -x +y}$$

$$f_{xx} = \frac{-1}{(1-x+y)^{2}}, f_{xy} = \frac{1}{(1-x+y)^{2}}, f_{yy} = \frac{-1}{(1-x+y)^{2}}$$

よって

$$\begin{array}{ll} f(1,1) = \log(1-1+1) = 0 & f_{x}(1,1) = \fr... ...xx}(1,1) = -1\\ f_{xy}(1,1) = 1 & f_{yy}(1,1) = -1 \end{array}$$

したがって，

$$f(x,y) = 0 - (x-1) + (y - 1) + \frac{1}{2}\left(-(x-1)^{2} + 2(x-1)(y - 1) - (y - 1)^{2}\right)$$

$$= -(x-1) + y-1 + \frac{1}{2}\left(-(x-1)^{2} + 2(x-1)(y-1) - (y-1)^{2}\right)$$

(d)

$$f_{x} = e^{2x+y} + 2xe^{2x+y},\ f_{y} = xe^{2x+y}$$

$$f_{xx} = 2e^{2x+y} + 2e^{2x+y} + 4xe^{2x+y},\ f_{xy} = e^{2x+y} + 2xe^{2x+y}, f_{yy} = xe^{2x+y}$$

よって

$$\begin{array}{ll} f(0,0) = 0 & f_{x}(0,0) = 1\\ f_{y}(0,0) =... ...0,0) = 2+2 = 4\\ f_{xy}(0,0) = 1 & f_{yy}(0,0) = 0 \end{array}$$

したがって，

$$f(x,y) = x + \frac{1}{2}\left(4x^{2} + 2xy\right)$$

2.

(a) $f(x,y) = 2x - x^{2} - y^{2}$より $f_{x} = 2 - 2x, f_{y} = -2y$. $f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = 2 -2x_{0} = 0, f_{y}(x_{0},y_{0}) = -2y_{0} = 0$$

これを $x_{0},y_{0}$について解くと $x_{0} = 1， y_{0} = 0$. 次に $f_{xx}(x,y) =-2, f_{xy}(x,y) = 0, f_{yy}(x,y) = -2$, $\Delta\mid_{(1,0)} = f_{xx}f_{yy} - (f_{xy})^2 = (-2)(-2) = 4 > 0$より $\Delta > 0, A = f_{xx}(1,0) = -2 < 0$となるので $f(1,0) = 2 -1 = 1$は極大値．

(b) $f(x,y) = x^{2} - 6y^{2} + y^{3}$より $f_{x} = 2x, f_{y} = -12y+ 3y^{2}$. $f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = 2x_{0} = 0, f_{y}(x_{0},y_{0}) = -12y_{0} + 3y_{0}^{2} = 3y_{0}(y_{0} - 4) = 0$$

これを $x_{0},y_{0}$について解くと $x_{0} = 0， y_{0} = 0,4$. 次に $f_{xx}(x,y) = 2, f_{xy}(x,y) = 0, f_{yy}(x,y) = -12 + 6y$, $\Delta\mid_{(0,0)} = f_{xx}f_{yy} - (f_{xy})^2 = 2(-12) = -24 < 0$より $f(0,0)$は鞍点．また， $\Delta\mid_{(0,4)} = f_{xx}f_{yy} - (f_{xy})^2 = 2(-12+24) = 24 > 0$, $A = f_{xx}(0,4) = 2 > 0$より $f(0,4) = -96 + 64 = -32$は極小値．

(c) $f(x,y) = x^{3} - 3x + y^{2}$より $f_{x} = 3x^{2} - 3, f_{y} = 2y$. $f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = 3x_{0}^{2} - 3 = 0, f_{y}(x_{0},y_{0}) = 2y_{0} = 0$$

これを $x_{0},y_{0}$について解くと $x_{0} = \pm 1， y_{0} = 0$. 次に $f_{xx}(x,y) = 6x, f_{xy}(x,y) = 0, f_{yy}(x,y) = 2$, $\Delta\mid_{(1,0)} = f_{xx}f_{yy} - (f_{xy})^2 = 12 > 0$より $\Delta > 0, A = f_{xx}(1,0) = 6 > 0$となるので， $f(1,0) = -2$は極小値．また， $\Delta\mid_{(-1,0)} = f_{xx}f_{yy} - (f_{xy})^2 = -6\cdot 2 = -12 < 0$より$f(-1,0)$は鞍点

(d) $f(x,y) = x^{2} + xy + y^{2} - 3x -3y$より $f_{x} = 2x + y -3, f_{y} = x + 2y - 3$. $f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = 2x_{0} + y_{0} - 3 = 0, f_{y}(x_{0},y_{0}) =x_{0} + 2y_{0} - 3= 0$$

これを $x_{0},y_{0}$について解くと $x_{0} = 1， y_{0} = 1$. 次に $f_{xx}(x,y) = 2, f_{xy}(x,y) = 1, f_{yy}(x,y) = 2$, $\Delta\mid_{(1,1)} = f_{xx}f_{yy} - (f_{xy})^2 = 2\cdot 2 - 1 = 3 > 0$より $\Delta > 0, A = f_{xx}(1,1) = 2 > 0$となるので， $f(1,1) = -1 + 1+ 1 -3- 3 = -3$は極小値．