6.9 Answer

6.9

1. Suppose that implicit function $y = g(x)$ determined by $f(x,y) = 0$ attains the extremum $y_{0} = g(x_{0})$ at $x = x_{0}$. Then

$$f(x_{0},y_{0}) = 0, f_{x}(x_{0},y_{0}) = 0, f_{y}(x_{0},y_{0}) \neq 0$$

Furthermore

$$\frac{d^2 y}{dx^2}\mid_{(x_{0},y_{0})} = -\frac{f_{xx}(x_{0},y_{0})}{f_{y}(x_{0},y_{0})} > 0 \ implies \ y_{0} = g(x_{0}) \ is the local minimum．$$

$$\frac{d^2 y}{dx^2}\mid_{(x_{0},y_{0})} = -\frac{f_{xx}(x_{0},y_{0})}{f_{y}(x_{0},y_{0})} < 0 \ implies \ y_{0} = g(x_{0}) \ is the local maximum．$$

(a) First we find $f(x,y)$ such that $f(x,y) = 0, f_{x}(x,y) = 0$.

$$f(x,y) = 8x^2 + 4xy + 5y^2 -36 = 0, f_{x}(x,y) = 16x + 4y = 0$$

implies $(\frac{\sqrt{2}}{2}, -2\sqrt{2}), (-\frac{\sqrt{2}}{2}, 2\sqrt{2})$． Next ，

$$f_{xx} = 16, \ f_{y} = 4x + 10y$$

Then calculate $\frac{d^2 y}{dx^2}$.

$$\frac{d^2 y}{dx^2}\mid_{(\frac{\sqrt{2}}{2}, -2\sqrt{2})} = -\frac{16}{2\sqrt{2} - 20\sqrt{2}} > 0$$

$$\frac{d^2 y}{dx^2}\mid_{(-\frac{\sqrt{2}}{2}, 2\sqrt{2})} = -\frac{16}{-2\sqrt{2} + 20\sqrt{2}} < 0$$

Thus for $x = \frac{\sqrt{2}}{2}$, $y = -2\sqrt{2}$ is the local minimum and for $x = -\frac{\sqrt{2}}{2}$, $y = 2\sqrt{2}$ is the local maximum．

(b)

First we find $f(x,y)$ so that $f(x,y) = 0, f_{x}(x,y) = 0$.

$$f(x,y) = x^2 y + x + y = 0, f_{x}(x,y) = 2xy + 1 = 0$$

Then $y = -\frac{1}{2x}$．Put this into $f(x,y) = 0$. Then

$$x^2 (-\frac{1}{2x}) + x - \frac{1}{2x} = \frac{x^2 -1}{2x} = 0$$

Thus $(1, -\frac{1}{2}), (-1, \frac{1}{2})$． Next，

$$f_{xx} = 2y, \ f_{y} = 2x$$

Then calculate $\frac{d^2 y}{dx^2}$.

$$\frac{d^2 y}{dx^2}\mid_{(1, -\frac{1}{2})} = -\frac{-1}{2} > 0$$

$$\frac{d^2 y}{dx^2}\mid_{(-1, \frac{1}{2})} = -\frac{1}{-2} > 0$$

Thus for $x = 1$, $y = -\frac{1}{2}$ is the local minimum and for $x = -1$, $y = \frac{1}{2}$ is also the local minimum．

(c)

First we find $(x,y)$ so that $f(x,y) = 0, f_{x}(x,y) = 0$．

$$f(x,y) = x^3 + y^3 - 6xy = 0, f_{x}(x,y) = 3x^2 - 6y = 0$$

Then $y = \frac{x^2}{2}$．Put this inot $f(x,y) = 0$. Then

$$x^3 + \frac{x^6}{8} - 3x^3 = \frac{x^3}{8}(x^3 - 16) = 0$$

Thus, $(0,0), (2\sqrt[3]{2}, 2\cdot2^{\frac{2}{3}})$． Next，

$$f_{xx} = 6x, \ f_{y} = -6$$

Then calculate $\frac{d^2 y}{dx^2}$.

$$\frac{d^2 y}{dx^2}\mid_{(0,0)} = -\frac{0}{-6} = 0$$

$$\frac{d^2 y}{dx^2}\mid_{(2\sqrt[3]{2}, 2\cdot2^{\frac{2}{3}})} = -\frac{6\cdot 2\sqrt[3]{2}}{-6} > 0$$

Thus for $x = 2\sqrt[3]{2}$, $y = 2\cdot2^{\frac{2}{3}}$ is the local minimum．

2. Suppose that at least one of $g_{x}(x_{0},y_{0})$ or $g_{y}(x_{0},y_{0})$ is not 0． $f(x,y)$ attains the extremun at $(x_{0},y_{0})$ under the condition $g(x,y) = 0$ if

$$F(x,y,\lambda) = f(x,y) - \lambda g(x,y)$$

and

$$F_{\lambda}(x,y) = 0, F_{x}(x,y) = 0, F_{y}(x,y) = 0$$

at $(x_{0},y_{0})$. Here, the points satisfying $g(x,y) = 0, g_{x}(x,y) = g_{y}(x,y) = 0$ are called singular points.

(a) Let $F(x,y,\lambda) = xy^3 - \lambda(x^2 + y^2 -1)$.

$$F_{\lambda} = 0 \ \ x^2 + y^2 -1 = 0$$ (1.9)

$$F_{x} = 0 \ \ y^3 - 2x\lambda = 0$$ (1.10)

$$F_{y} = 0 \ \ 3xy^2 - 2y\lambda = 0$$ (1.11)

For $y=0$, the equation (1.9) implies， $x = \pm 1$ and the equation (1.10) implies $\lambda = 0$． For $y \neq 0$，the equation (1.11) implies $2\lambda = 3xy$．Now put this into the equation (1.10). Then $y^3 - x(3xy) = y(y^2 - 3x^2) = 0$ implies $y^2 = 3x^2$．Now put this into the equation (1.9). Then $x^2 + 3x^2 - 1 = 0$ implies $x = \pm \frac{1}{2}$．Thus， $y = \pm \frac{\sqrt{3}}{2}$．Therefore the solution to the equations (1.9),(1.10),(1.11) are

$$(x,y) = (1,0),(-1,0),(\frac{1}{2},\frac{\sqrt{3}}{2}),(-\frac{1}{... ...rt{3}}{2}),(\frac{1}{2},-\frac{\sqrt{3}}{2}),(-\frac{1}{2},-\frac{\sqrt{3}}{2})$$

Then the maximum value of $xy^3$ is

$$0,0,\frac{3\sqrt{3}}{16},-\frac{3\sqrt{3}}{16},-\frac{3\sqrt{3}}{16},\frac{3\sqrt{3}}{16}$$

On the other hand, $g(x,y) = 0$ is bounded closed region and $f$ is continuous on this region. Thus it takes the maximum and the minimum. Thus, the maximum value is

$$f(\frac{1}{2},\frac{\sqrt{3}}{2}) = f(-\frac{1}{2},-\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{16}，$$

The minimum value is

$$f(-\frac{1}{2},\frac{\sqrt{3}}{2}) = f(\frac{1}{2},-\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{16}．$$

(b) Let $F(x,y,\lambda) = x^2 + y^2 - \lambda(x^3 + y^3 - 6xy)$. Then

$$F_{\lambda} = 0 \ + y^3 - 6xy = 0$$ (1.12)

$$F_{x} = 0 \ \ 2x - \lambda(3x^2 - 6y) = 0$$ (1.13)

$$F_{y} = 0 \ \ 2y - \lambda(3y^2 - 6x) = 0$$ (1.14)

From the equations (1.12),(1.13),

$$\lambda = \frac{2x}{3x^2 - 6y} = \frac{2y}{3y^2 - 6x}$$

$$6xy^2 - 12x^2 = 6x^2 y -12y^2$$

$$(x-y)(xy + 2(x+y)) = 0$$

Thus，$x = y$ or $y = -\frac{2x}{x + 2}$． $F_{\lambda} = 0$ implies $x^3 + x^3 - 6x^2 = 2x^2(x - 3) = 0$．Therefore $(x = 0, y= 0), (x = 3, y = 3)$． On the other hand, The part of $g(x,y) = 0$ in the 1st quadrant with the origin is the bounded closed curve and $f(x,y)$ is continuous on the curve. Thus, it takes the maximum and the minimum. Therefore, the maximimun value is

$$f(3,3) = 18.$$

Also，if $(x,y) \neq (0,0)$, then $f(x,y) > 0$. Thus $f(0,0) = 0$ is the local minimum and also the minimum．

(c) Let $F(x,y,\lambda) = xy - \lambda(x^2 - xy + y^2 - 1)$.

$$F_{\lambda} = 0 \ より \ x^2 - xy + y^2 - 1 = 0$$ (1.15)

$$F_{x} = 0 \ より \ y - \lambda(2x - y) = 0$$ (1.16)

$$F_{y} = 0 \ より \ x - \lambda(2y - x) = 0$$ (1.17)

By the equations (1.16),(1.17),

$$\lambda = \frac{y}{2x-y} = \frac{x}{2y-x}$$

$$2y^2 - xy = 2x^2 - xy$$

$$x^2 = y^2$$

Thus，for $x = \pm y$．$x = y$, $F_{\lambda} = 0$ implies $x^2 - x^2 + x^2 - 1 = x^2 - 1 = 0$．Thus， $(x = 1, y= 1),(x = -1, y = -1)$．$x=-y$のとき $x^2 + x^2 + x^2 - 1 = 3x^2 - 1 = 0$．Therefore， $(x = \frac{\sqrt{3}}{3},y=-\frac{\sqrt{3}}{3}),(x = -\frac{sqrt{3}}{3}, y = \frac{\sqrt{3}}{3})$． Then，The value of $f(x,y) = xy$ is，

$$1,1,-\frac{1}{3},-\frac{1}{3}$$

ther hand, On the ， $g(x,y) = 0$ is bounded closed region and $f$ is continuous on this region. Thus, $f$ takes the maximum and the minimum. The maximum value is

$$f(1,1) = f(-1,-1) = 1，$$

The minimum value is

$$f(\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}) = f(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}) = -\frac{1}{3}．$$

3. Let $g(x,y) = 2x + 3y - 12 = 0$. Then $F(x,y,\lambda) = xy - \lambda(2x + 3y - 12)$.

$$F_{\lambda} = 0 \ より \ 2x + 3y -12 = 0$$ (1.18)

$$F_{x} = 0 \ より \ y - 2\lambda = 0$$ (1.19)

$$F_{y} = 0 \ より \ x - 3\lambda = 0$$ (1.20)

From the equations (1.19),(1.20), we have $y=2\lambda, x = 3\lambda$．Put this into the equation (1.18). Then

$$6\lambda + 6\lambda -12 = 12\lambda -12 = 0$$

Thus， $\lambda = 1$ and $(x = 3, y = 2)$．On the other hand, $g(x,y) = 0$ is bounded closed region and $f$ is continuous on this region. Thus, $f$ attains the maximum and minimum.

$$f(3,2) = 6.$$

4.

Let $g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0$. $F(x,y,z,\lambda) = x^2 + 2y^2 + 3z^2 - \lambda(x^2 + y^2 + z^2 - 1)$.

$$F_{\lambda} = 0 \ \ x^2 + y^2 + z^2 -1 = 0$$ (1.21)

$$F_{x} = 0 \ \ 2x - 2x\lambda = 0$$ (1.22)

$$F_{y} = 0 \ \ 4y - 2y\lambda = 0$$ (1.23)

$$F_{z} = 0 \ implies \ 6z - 2z\lambda = 0$$ (1.24)

By the equations (1.22),(1.23),(1.24), $2x(1-\lambda)=0$, $2y(2 - \lambda) = 0$, $2z(3-\lambda) = 0$．Here，for $x \neq 0$, $\lambda = 1$implies $y = z = 0$．Put this into (1.21).

$$x^2 - 1 = 0 \$$   implies$$\ x = \pm 1$$

Thus， $(1,0,0), (-1,0,0)$．For $y \neq 0$， $\lambda = 2$ implies $x = z = 0$．Put this into the equation (1.21).

$$y^2 - 1 = 0 \ より \ y = \pm 1$$

Thus， $(0,1,0), (0,-1,0)$．For $z \neq 0$， $\lambda = 3$ implies $x = y = 0$．Put this intp the equation (1.21). Then

$$z^2 - 1 = 0 \$$   implies$$\ z = \pm 1$$

Thus， $(0,0,1), (0,0,-1)$．From this, we fin the value of the equation (1.21).

$$f(1,0,0) = f(-1,0,0) = 1, f(0,1,0) = f(0,-1,0) = 2, f(0,0,1) = f(0,0,-1) = 3$$

On the other hand， $g(x,y,z) = 0$ is bounded closed region and $f$ is continuous on this region. Thus, $f$ attains the maximum and minimum. The maximun value is

$$f(0,0,1) = f(0,0,-1) = 3.$$

The minimum value is

$$f(1,0,0) = f(-1,0,0) = 1.$$