6.7 解答

6.7

1.

(a) $f(x,y) = 2x^2 + y^2 - xy -7y$より $f_{x} = 4x - y, f_{y} = 2y - x - 7$. $f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = 4x_{0} - y_{0} = 0, f_{y}(x_{0},y_{0}) = 2y_{0} - x_{0} - 7 = 0$$

これを $x_{0},y_{0}$について解くと $x_{0} = 1， y_{0} = 4$. 次に $f_{xx}(x,y) = 4, f_{xy}(x,y) = -1, f_{yy}(x,y) = 2$, $\Delta = f_{xx}f_{yy} - (f_{xy})^2 = 8 -1 = 7 > 0$より $\Delta > 0, A = f_{xx}(1,4) = 4 > 0$となるので $f(1,4) = 2 + 16 - 4 - 28 = -14$は極小値．

(b)

$f(x,y) = \frac{x}{y^2} + xy$より $f_{x} = \frac{1}{y^2} + y, f_{y} = -\frac{2x}{y^2} + x$. $f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = \frac{1}{y_{0}^2} + y_{0} = 0, f_{y}(x_{0},y_{0}) = -\frac{2x_{0}}{y_{0}^3} + x_{0} = 0$$

これを $x_{0},y_{0}$について解く． $\frac{1}{y_{0}^2} + y_{0} = 0$より $\frac{1 + y_{0}^3}{y_{0}^2} = 0$．よって $y_{0} = -1$．また， $-\frac{2x_{0}}{y_{0}^3} + x_{0} = x_{0}(\frac{-2}{y_{0}^3} + 1) = 0$より$x_{0} = 0$．

次に $f_{xx}(x,y) = 0, f_{xy}(x,y) = -\frac{2}{y^3} + 1, f_{yy}(x,y) = \frac{6x}{y^4}$． $(0,1)$では

$$\Delta = f_{xx}(0,1)f_{yy}(0,1) - (f_{xy}(0,1))^2 = -9 < 0$$

したがって，極値なし．

(c)

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

$$f_{x}(x_{0},y_{0}) = \frac{1}{y_{0}^2} + y_{0} = 0$$ (10.1)

$$f_{y}(x_{0},y_{0}) = 3y_{0}^2 - 3x_{0} = 0$$ (10.2)

これを $x_{0},y_{0}$について解く．式(10.1)より $y_{0} = x_{0}^2$を式(10.2)に代入すると，

$$3x_{0}^4 - 3x_{0} = 3x_{0}(x_{0}^3 - 1) = 0$$

よって， $x_{0} = 0, 1$．したがって， $(x_{0}=0,y_{0}=0), (x_{0}=1, y_{0} =1)$． 次に $f_{xx}(x,y) = 6x, f_{xy}(x,y) = -3, f_{yy}(x,y) = 6y$． $(0,0)$では

$$\Delta = f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2 = -9 < 0$$

したがって，$f(0,0)$は極値でない． $(1,1)$では

$$\Delta = f_{xx}(1,1)f_{yy}(1,1) - (f_{xy}(1,1))^2 = 6\cdot6 - (-3)^2 = 27 > 0 , A = f_{xx}(1,1) = 6 > 0$$

したがって， $f(1,1) = 1 + 1 - 3 = -1$は極小値．

(d)

$f(x,y) = (x^2 + y^2)^2ー2(x^2 - y^2)$より

$$f_{x} = 2(x^2 + y^2)(2x) - 4x = 4x(x^2 + y^2 -1)$$

$$f_{y} = 4y(x^2 + y^2) + 4y = 4y(x^2 + y^2 + 1)$$

$f(x,y)$が $(x_{0},y_{0})$で極値をとるならば

$$f_{x}(x_{0},y_{0}) = 4x_{0}(x_{0}^2 + y_{0}^2 -1) = 0$$ (10.3)

$$f_{y}(x_{0},y_{0}) = 4y_{0}(x_{0}^2 + y_{0}^2 + 1) = 0$$ (10.4)

これを $x_{0},y_{0}$について解く．式(10.4)より$y_{0} = 0$となり，これを式(10.3)に代入すると，

$$4x_{0}(x_{0}^2 - 1) = 0$$

よって， $x_{0} = 0, \pm1$．したがって， $(x_{0}=0,y_{0}=0), (x_{0}=1, y_{0} =0), (x_{0}=-1, y_{0} = 0)$． 次に

$$f_{xx}(x,y) = 4(x^2 + y^2 -1) + 8x^2 = 4(3x^2 + y^2 -1)$$

$$f_{xy}(x,y) = 8xy$$

$$f_{yy}(x,y) = 4(x^2 + y^2 + 1) + 8y^2 = 4(x^2 + 3y^2 + 1)$$

$(0,0)$では

$$\Delta = f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2 = (-4)(4) - 0 = -16 < 0$$

したがって，$f(0,0)$は極値でない． $(1,0)$では

$$\Delta = f_{xx}(1,0)f_{yy}(1,0) - (f_{xy}(1,0))^2 = 8\cdot 8 - 0 = 64 > 0 , A = f_{xx}(1,0) = 8 > 0$$

したがって， $f(1,0) = -1$は極小値． $(-1,0)$では

$$\Delta = f_{xx}(-1,0)f_{yy}(-1,0) - (f_{xy}(-1,0))^2 = 8\cdot 8 - 0 = 64 > 0 , A = f_{xx}(-1,0) = 8 > 0$$

したがって， $f(-1,0) = -1$は極小値．

2. 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 $x_{0} = 0, y_{0}=0, h = x, k = y$とおくと

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

$$f_{x} = e^{x}\cos{y}, f_{y} = -e^{x}\sin{y}$$

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

よって

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

したがって，

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

(b) Taylorの定理で $x_{0}=2, y_{0} =1, x = x_{0} + h, y = k_{0} + k$とおくと

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

$$f_{x} = \frac{1}{x + y^2}, f_{y} = \frac{2y}{x + y^2}$$

$$f_{xx} = \frac{-1}{(x + y^2)^2}, f_{xy} = -\frac{-2y}{(x+y^2)^2}, f_{yy} = \frac{2(x+y^2)-4y^2}{(x+y^2)^2}$$

よって

$$f_{x}(2,1) = \frac{1}{3}, f_{y}(2,1) = \frac{2}{3}, f_{xx}(2,1) = \frac{-1}{9}, f_{xy}(2,1) = \frac{-2}{9}, f_{yy}(2,1) = \frac{2}{9}$$

したがって，

$$f(x,y) = \log{3} + \frac{1}{3}(x-2) + \frac{2}{3}(y-1) + \frac{1}{2}(-\frac{1}{9}(x-2)^2 -\frac{4}{9}(x-2)(y-1) + \frac{2}{9}(y-1)^2$$