6.7 Answer

6.7

1.

(a) Since $f(x,y) = 2x^2 + y^2 - xy -7y$, $f_{x} = 4x - y, f_{y} = 2y - x - 7$. If $f(x,y)$ attains the extreme at $(x_{0},y_{0})$

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

Solve this for $x_{0},y_{0}$, we have $x_{0} = 1, y_{0} = 4$. Next for $f_{xx}(x,y) = 4, f_{xy}(x,y) = -1, f_{yy}(x,y) = 2$, $\Delta = f_{xx}f_{yy} - (f_{xy})^2 = 8 -1 = 7 > 0$, we have $\Delta > 0, A = f_{xx}(1,4) = 4 > 0$. Thus $f(1,4) = 2 + 16 - 4 - 28 = -14$ is the local minimum．

(b)

Since $f(x,y) = \frac{x}{y^2} + xy$, we have $f_{x} = \frac{1}{y^2} + y, f_{y} = -\frac{2x}{y^2} + x$. If $f(x,y)$ attains the extreme at $(x_{0},y_{0})$, then

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

Solve this for $x_{0},y_{0}$．Then $\frac{1}{y_{0}^2} + y_{0} = 0$ implies $\frac{1 + y_{0}^3}{y_{0}^2} = 0$．Thus, $y_{0} = -1$．Also, $-\frac{2x_{0}}{y_{0}^3} + x_{0} = x_{0}(\frac{-2}{y_{0}^3} + 1) = 0$ implies $x_{0} = 0$．

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

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

Therefore, no extremum.．

(c)

Since $f(x,y) = x^3 + y^3 - 3xy$, $f_{x} = 3x_{0}^2 - 3y_{0}, f_{y} = -\frac{2x}{y^2} + x$. If $f(x,y)$ attains the extremum at $(x_{0},y_{0})$, then

$$f_{x}(x_{0},y_{0}) = \frac{1}{y_{0}^2} + y_{0} = 0$$ (1.1)

$$f_{y}(x_{0},y_{0}) = 3y_{0}^2 - 3x_{0} = 0$$ (1.2)

Solve this for $x_{0},y_{0}$. Put $y_{0} = x_{0}^2$. Then

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

Thus， $x_{0} = 0, 1$．Therefore， $(x_{0}=0,y_{0}=0), (x_{0}=1, y_{0} =1)$． Next $f_{xx}(x,y) = 6x, f_{xy}(x,y) = -3, f_{yy}(x,y) = 6y$． At $(0,0)$.

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

Thus，$f(0,0)$ is not extremum． At $(1,1)$, we have

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

Therefore， $f(1,1) = 1 + 1 - 3 = -1$ is the local minimum．

(d)

Since $f(x,y) = (x^2 + y^2)^2ー2(x^2 - y^2)$, we have

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

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

If $f(x,y)$ attains the extremum at $(x_{0},y_{0})$, then

$$f_{x}(x_{0},y_{0}) = 4x_{0}(x_{0}^2 + y_{0}^2 -1) = 0$$ (1.3)

$$f_{y}(x_{0},y_{0}) = 4y_{0}(x_{0}^2 + y_{0}^2 + 1) = 0$$ (1.4)

Solve this for $x_{0},y_{0}$．Then

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

Thus， $x_{0} = 0, \pm1$．Therefore， $(x_{0}=0,y_{0}=0), (x_{0}=1, y_{0} =0), (x_{0}=-1, y_{0} = 0)$． 次に

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

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

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

At $(0,0)$, we have

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

Thus，$f(0,0)$ is not extremum． $(1,0)$では

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

Thus， $f(1,0) = -1$ is local minimum． At $(-1,0)$, we have

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

Thus， $f(-1,0) = -1$ is th elocal minimum．

2. By the Taylor's theorem, we have

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

(a) By the Taylor's theorem, let $x_{0} = 0, y_{0}=0, h = x, k = y$. Then

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

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

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

Thus

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

Therefore,，

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

(b) By the Taylor's theorem, we let $x_{0}=2, y_{0} =1, x = x_{0} + h, y = k_{0} + k$. Then

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

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

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

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，

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