Lagrange Multipliers

$f = 2x + y$, $\ g = x^2 + 4y^2 = 17$.

$2 = 2\lambda x$, $\ 1 = 8\lambda y$, $\ x^2 + 4y^2 = 17$.

$\lambda \neq 0$, so $x = \dfrac{1}{\lambda}$ and $y = \dfrac{1}{8\lambda}$: $x = 8y$. Then $64y^2 + 4y^2 = 17$, $y = \pm\tfrac12$, $x = \pm 4$.

$f\left(4, \tfrac12\right) = \tfrac{17}{2}$ and $f\left(-4, -\tfrac12\right) = -\tfrac{17}{2}$.

$f = x^2y$, $\ g = x^2 + y^2 = 3$.

$2xy = 2\lambda x$, $\ x^2 = 2\lambda y$, $\ x^2 + y^2 = 3$.

First equation: $x(y - \lambda) = 0$. Case $x = 0$: $y = \pm\sqrt3$. Case $\lambda = y$: $x^2 = 2y^2$, so $3y^2 = 3$, $y = \pm 1$, $x = \pm\sqrt2$.

$f(0, \pm\sqrt3) = 0$, $\ f(\pm\sqrt2, 1) = 2$, $\ f(\pm\sqrt2, -1) = -2$.

$f = x^2 + 4y^2$, $\ g = xy = 2$.

$2x = \lambda y$, $\ 8y = \lambda x$, $\ xy = 2$.

Multiply the first by $x$ and the second by $y$: $2x^2 = \lambda xy = 8y^2$, so $x = \pm 2y$. Since $xy = 2 > 0$, $x = 2y$: $2y^2 = 2$, $y = \pm 1$.

$f(2, 1) = f(-2, -1) = 8$. Along $xy = 2$, $f \to \infty$ as $x \to \infty$, so $8$ is the min and there is no max.

$2 = 2\lambda x$, $\ -1 = 2\lambda y$, $\ 2 = 2\lambda z$, $\ x^2 + y^2 + z^2 = 9$.

$x = \dfrac{1}{\lambda}$, $\ y = -\dfrac{1}{2\lambda}$, $\ z = \dfrac{1}{\lambda}$.

$\dfrac{1 + \tfrac14 + 1}{\lambda^2} = 9$ gives $\lambda = \pm\tfrac12$: the points $\pm(2, -1, 2)$.

$f(2, -1, 2) = 4 + 1 + 4 = 9$ and $f(-2, 1, -2) = -9$.

$yz = 2\lambda x$, $\ xz = 4\lambda y$, $\ xy = 8\lambda z$, $\ x^2 + 2y^2 + 4z^2 = 12$.

Multiply by $x$, $y$, $z$: $xyz = 2\lambda x^2 = 4\lambda y^2 = 8\lambda z^2$. (If $\lambda = 0$, two of $x, y, z$ are $0$ and $f = 0$.)

So $x^2 = 2y^2 = 4z^2$, and the constraint gives $3x^2 = 12$: $x^2 = 4$, $y^2 = 2$, $z^2 = 1$.

$f = \pm 2 \cdot \sqrt2 \cdot 1 = \pm 2\sqrt2$, depending on the signs.

$D = x^2 + y^2$.

$2x = \lambda(2x + y)$, $\ 2y = \lambda(x + 2y)$, $\ x^2 + xy + y^2 = 3$.

Eliminate $\lambda$: $2x(x + 2y) = 2y(2x + y)$, so $x^2 = y^2$. $y = x$: $3x^2 = 3$, points $\pm(1, 1)$. $y = -x$: $x^2 = 3$, points $\pm(\sqrt3, -\sqrt3)$.

$D(\pm(1, 1)) = 2$ and $D(\pm(\sqrt3, -\sqrt3)) = 6$.

$D = (x - 3)^2 + y^2 + (z - 4)^2$.

$2(x - 3) = 2\lambda x$, $\ 2y = 2\lambda y$, $\ 2(z - 4) = 2\lambda z$, $\ x^2 + y^2 + z^2 = 4$.

$\lambda = 1$ would give $-3 = 0$, so $\lambda \neq 1$ and $y = 0$. Then $x = \dfrac{3}{1 - \lambda}$, $z = \dfrac{4}{1 - \lambda}$, so $(x, z)$ is a multiple of $(3, 4)$. With $x^2 + z^2 = 4$, that gives $(x, z) = \pm\left(\tfrac65, \tfrac85\right)$.

$D\left(\tfrac65, 0, \tfrac85\right) = 9$ and $D\left(-\tfrac65, 0, -\tfrac85\right) = 49$.

$D = x^2 + y^2 + z^2$, with $g = z^2 - xy = 4$.

$2x = -\lambda y$, $\ 2y = -\lambda x$, $\ 2z = 2\lambda z$, $\ z^2 - xy = 4$.

$z(1 - \lambda) = 0$. Case $\lambda = 1$: $2x = -y$ and $2y = -x$ force $x = y = 0$, so $z = \pm 2$. Case $z = 0$: $xy = -4$ and $2x^2 = -\lambda xy = 2y^2$, so $y = -x$, $x = \pm 2$: points $\pm(2, -2, 0)$.

$D(0, 0, \pm 2) = 4$ and $D(\pm(2, -2, 0)) = 8$.

Base $x$ by $y$, height $z$. Cost $C = 5(2xy) + 2(2xz + 2yz) = 10xy + 4xz + 4yz$.

$g = xyz = 20$.

$10y + 4z = \lambda yz$, $\ 10x + 4z = \lambda xz$, $\ 4x + 4y = \lambda xy$. Multiply the first by $x$ and the second by $y$: $10xy + 4xz = 10xy + 4yz$, so $x = y$. Then $8x = \lambda x^2$, $\lambda = \tfrac{8}{x}$, and $10x + 4z = 8z$ gives $z = \tfrac52 x$. So $\tfrac52 x^3 = 20$, $x = 2$.

$x = y = 2\text{ m}$, $z = 5\text{ m}$; cost $40 + 40 + 40 = 120$ dollars.

Corner $(x, y, z)$ with $x, y, z > 0$; $V = 8xyz$.

$g = \dfrac{x^2}{4} + \dfrac{y^2}{9} + z^2 = 1$.

$8yz = \dfrac{\lambda x}{2}$, $\ 8xz = \dfrac{2\lambda y}{9}$, $\ 8xy = 2\lambda z$. Multiply them by $x$, $y$, $z$: $8xyz = 2\lambda\dfrac{x^2}{4} = 2\lambda\dfrac{y^2}{9} = 2\lambda z^2$, so each term of $g$ is $\tfrac13$: $x = \dfrac{2}{\sqrt3}$, $y = \sqrt3$, $z = \dfrac{1}{\sqrt3}$.

Sides $\dfrac{4}{\sqrt3} \times 2\sqrt3 \times \dfrac{2}{\sqrt3}$; $V = \dfrac{16}{\sqrt3} = \dfrac{16\sqrt3}{3}$.

$f = xy^2z$.

$g = x + y + z = 12$.

$y^2z = \lambda$, $\ 2xyz = \lambda$, $\ xy^2 = \lambda$. From the first and third, $z = x$; from the first and second, $y = 2x$. So $4x = 12$, $x = 3$.

$x = 3$, $y = 6$, $z = 3$: $f = 3 \cdot 36 \cdot 3 = 324$.

$f = z$; $\ g = x^2 + y^2 = 2$; $\ h = x + y + z = 1$.

$0 = 2\lambda x + \mu$, $\ 0 = 2\lambda y + \mu$, $\ 1 = \mu$.

$\mu = 1$, so $2\lambda x = 2\lambda y = -1$: $\lambda \neq 0$ and $x = y$. Then $2x^2 = 2$, $x = \pm 1$, and $z = 1 - 2x$.

$(1, 1, -1)$ gives $z = -1$; $\ (-1, -1, 3)$ gives $z = 3$.

$D = x^2 + y^2 + z^2$; $\ g = x - y + z = 1$; $\ h = 2x + y + 3z = 10$.

$2x = \lambda + 2\mu$, $\ 2y = -\lambda + \mu$, $\ 2z = \lambda + 3\mu$.

Substitute into the constraints: $3\lambda + 4\mu = 2$ and $2\lambda + 7\mu = 10$, so $\mu = 2$, $\lambda = -2$. Then $(x, y, z) = (1, 2, 2)$.

$D = 1 + 4 + 4 = 9$. A line has one closest point, so this is the min.