関数 $f(x, y)$ を条件 $g(x, y) = 0$ のもとで、ラグランジュの未定乗数法を用いて最大値と最小値を求める問題です。具体的には、以下の6つのケースについて解きます。 1) $f(x, y) = x + y, \quad g(x, y) = x^2 + y^2 - 1$ 2) $f(x, y) = 2x + y, \quad g(x, y) = x^2 + y^2 - 1$ 3) $f(x, y) = xy, \quad g(x, y) = x^2 + y^2 - 1$ 4) $f(x, y) = \frac{x}{2} + y, \quad g(x, y) = (\frac{x}{2})^2 + y^2 - 1$ 5) $f(x, y) = x + y, \quad g(x, y) = (\frac{x}{2})^2 + y^2 - 1$ 6) $f(x, y) = \frac{xy}{2}, \quad g(x, y) = (\frac{x}{2})^2 + y^2 - 1$

1. 問題の内容

関数 $f(x, y)$ を条件 $g(x, y) = 0$ のもとで、ラグランジュの未定乗数法を用いて最大値と最小値を求める問題です。具体的には、以下の6つのケースについて解きます。

1) $f(x, y) = x + y, \quad g(x, y) = x^2 + y^2 - 1$

2) $f(x, y) = 2x + y, \quad g(x, y) = x^2 + y^2 - 1$

3) $f(x, y) = xy, \quad g(x, y) = x^2 + y^2 - 1$

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

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

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

2. 解き方の手順

ラグランジュの未定乗数法では、ラグランジアン $L(x, y, \lambda) = f(x, y) - \lambda g(x, y)$ を定義し、その偏微分がすべて0になる点を求めます。

$\frac{\partial L}{\partial x} = 0$

$\frac{\partial L}{\partial y} = 0$

$\frac{\partial L}{\partial \lambda} = -g(x, y) = 0$

これらの連立方程式を解き、$x, y, \lambda$ を求めます。求めた $x, y$ を $f(x, y)$ に代入することで、極値を求めます。極値の中から最大値と最小値を決定します。

**ケース1)**

$f(x, y) = x + y, \quad g(x, y) = x^2 + y^2 - 1$

$L(x, y, \lambda) = x + y - \lambda(x^2 + y^2 - 1)$

$\frac{\partial L}{\partial x} = 1 - 2\lambda x = 0$

$\frac{\partial L}{\partial y} = 1 - 2\lambda y = 0$

$\frac{\partial L}{\partial \lambda} = -(x^2 + y^2 - 1) = 0$

$x = y = \frac{1}{2\lambda}$

$x^2 + y^2 = 1 \implies 2x^2 = 1 \implies x = \pm \frac{1}{\sqrt{2}}$

したがって、$x = y = \frac{1}{\sqrt{2}}$ または $x = y = -\frac{1}{\sqrt{2}}$

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

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

**ケース2)**

$f(x, y) = 2x + y, \quad g(x, y) = x^2 + y^2 - 1$

$L(x, y, \lambda) = 2x + y - \lambda(x^2 + y^2 - 1)$

$\frac{\partial L}{\partial x} = 2 - 2\lambda x = 0$

$\frac{\partial L}{\partial y} = 1 - 2\lambda y = 0$

$\frac{\partial L}{\partial \lambda} = -(x^2 + y^2 - 1) = 0$

$x = \frac{1}{\lambda}, y = \frac{1}{2\lambda}$

$x^2 + y^2 = 1 \implies \frac{1}{\lambda^2} + \frac{1}{4\lambda^2} = 1 \implies \frac{5}{4\lambda^2} = 1 \implies \lambda = \pm \frac{\sqrt{5}}{2}$

$x = \pm \frac{2}{\sqrt{5}}, y = \pm \frac{1}{\sqrt{5}}$ （複号同順）

$f(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}) = \frac{5}{\sqrt{5}} = \sqrt{5}$

$f(-\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}) = -\frac{5}{\sqrt{5}} = -\sqrt{5}$

**ケース3)**

$f(x, y) = xy, \quad g(x, y) = x^2 + y^2 - 1$

$L(x, y, \lambda) = xy - \lambda(x^2 + y^2 - 1)$

$\frac{\partial L}{\partial x} = y - 2\lambda x = 0$

$\frac{\partial L}{\partial y} = x - 2\lambda y = 0$

$\frac{\partial L}{\partial \lambda} = -(x^2 + y^2 - 1) = 0$

$y = 2\lambda x, x = 2\lambda y \implies x = 4\lambda^2 x$

$x=0$ or $4\lambda^2 = 1 \implies \lambda = \pm \frac{1}{2}$

If $x = 0$, $y = 0$, but $x^2 + y^2 = 1$.

If $\lambda = \frac{1}{2}, y = x$, so $x^2 + x^2 = 1 \implies x = \pm \frac{1}{\sqrt{2}}$. $y=x$.

If $\lambda = -\frac{1}{2}, y = -x$, so $x^2 + (-x)^2 = 1 \implies x = \pm \frac{1}{\sqrt{2}}$. $y=-x$.

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

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

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

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

**ケース4)**

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

$L(x, y, \lambda) = \frac{x}{2} + y - \lambda(\frac{x^2}{4} + y^2 - 1)$

$\frac{\partial L}{\partial x} = \frac{1}{2} - \frac{\lambda x}{2} = 0$

$\frac{\partial L}{\partial y} = 1 - 2\lambda y = 0$

$\frac{\partial L}{\partial \lambda} = -(\frac{x^2}{4} + y^2 - 1) = 0$

$x = \frac{1}{\lambda}, y = \frac{1}{2\lambda}$

$\frac{1}{4\lambda^2} + \frac{1}{4\lambda^2} = 1 \implies \frac{2}{4\lambda^2} = 1 \implies \lambda = \pm \frac{1}{\sqrt{2}}$

$x = \pm \sqrt{2}, y = \pm \frac{\sqrt{2}}{2}$ (複号同順)

$f(\sqrt{2}, \frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

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

**ケース5)**

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

$L(x, y, \lambda) = x + y - \lambda(\frac{x^2}{4} + y^2 - 1)$

$\frac{\partial L}{\partial x} = 1 - \frac{\lambda x}{2} = 0$

$\frac{\partial L}{\partial y} = 1 - 2\lambda y = 0$

$\frac{\partial L}{\partial \lambda} = -(\frac{x^2}{4} + y^2 - 1) = 0$

$x = \frac{2}{\lambda}, y = \frac{1}{2\lambda}$

$\frac{1}{\lambda^2} + \frac{1}{4\lambda^2} = 1 \implies \frac{5}{4\lambda^2} = 1 \implies \lambda = \pm \frac{\sqrt{5}}{2}$

$x = \pm \frac{4}{\sqrt{5}}, y = \pm \frac{1}{\sqrt{5}}$ (複号同順)

$f(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}) = \frac{5}{\sqrt{5}} = \sqrt{5}$

$f(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}) = -\frac{5}{\sqrt{5}} = -\sqrt{5}$

**ケース6)**

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

$L(x, y, \lambda) = \frac{xy}{2} - \lambda(\frac{x^2}{4} + y^2 - 1)$

$\frac{\partial L}{\partial x} = \frac{y}{2} - \frac{\lambda x}{2} = 0$

$\frac{\partial L}{\partial y} = \frac{x}{2} - 2\lambda y = 0$

$\frac{\partial L}{\partial \lambda} = -(\frac{x^2}{4} + y^2 - 1) = 0$

$y = \lambda x, x = 4\lambda y \implies x = 4\lambda^2 x$

$x=0$ or $4\lambda^2 = 1 \implies \lambda = \pm \frac{1}{2}$

If $x = 0$, $y = 0$, but this doesn't satisfy $g(x,y)=0$

If $\lambda = \frac{1}{2}, y = \frac{x}{2}$, so $\frac{x^2}{4} + \frac{x^2}{4} = 1 \implies x^2 = 2 \implies x = \pm \sqrt{2}$. $y = \pm \frac{\sqrt{2}}{2}$.

If $\lambda = -\frac{1}{2}, y = -\frac{x}{2}$, so $\frac{x^2}{4} + \frac{x^2}{4} = 1 \implies x^2 = 2 \implies x = \pm \sqrt{2}$. $y = \mp \frac{\sqrt{2}}{2}$.

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

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

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

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

3. 最終的な答え

**ケース1)** 最大値: $\sqrt{2}$, 最小値: $-\sqrt{2}$

**ケース2)** 最大値: $\sqrt{5}$, 最小値: $-\sqrt{5}$

**ケース3)** 最大値: $\frac{1}{2}$, 最小値: $-\frac{1}{2}$

**ケース4)** 最大値: $\sqrt{2}$, 最小値: $-\sqrt{2}$

**ケース5)** 最大値: $\sqrt{5}$, 最小値: $-\sqrt{5}$

**ケース6)** 最大値: $\frac{1}{2}$, 最小値: $-\frac{1}{2}$