We are asked to solve several optimization problems using Lagrange multipliers. 1. Find the minimum of $f(x, y) = x^2 + y^2$ subject to the constraint $g(x, y) = xy - 3 = 0$.

1. Problem Description

We are asked to solve several optimization problems using Lagrange multipliers.

1. Find the minimum of $f(x, y) = x^2 + y^2$ subject to the constraint $g(x, y) = xy - 3 = 0$.

2. Find the maximum of $f(x, y) = xy$ subject to the constraint $g(x, y) = 4x^2 + 9y^2 - 36 = 0$.

3. Find the maximum of $f(x, y) = 4x^2 - 4xy + y^2$ subject to the constraint $x^2 + y^2 = 1$.

4. Find the minimum of $f(x, y) = x^2 + 4xy + y^2$ subject to the constraint $x - y - 6 = 0$.

5. Find the minimum of $f(x, y, z) = x^2 + y^2 + z^2$ subject to the constraint $x + 3y - 2z = 12$.

6. Find the minimum of $f(x, y, z) = 4x - 2y + 3z$ subject to the constraint $2x^2 + y^2 - 3z = 0$.

7. What are the dimensions of the rectangular box, open at the top, that has maximum volume when the surface area is 48?

2. Solution Steps

1. $f(x, y) = x^2 + y^2$, $g(x, y) = xy - 3 = 0$

We use the method of Lagrange multipliers.

$\nabla f = (2x, 2y)$

$\nabla g = (y, x)$

$2x = \lambda y$

$2y = \lambda x$

$xy = 3$

From the first two equations, we have $\lambda = \frac{2x}{y} = \frac{2y}{x}$. Then $2x^2 = 2y^2$, so $x^2 = y^2$ and $x = \pm y$.

Since $xy = 3$, $x$ and $y$ have the same sign, so $x = y$.

$x^2 = 3$, so $x = \pm \sqrt{3}$. Since $x=y$, we have two points: $(\sqrt{3}, \sqrt{3})$ and $(-\sqrt{3}, -\sqrt{3})$.

$f(\sqrt{3}, \sqrt{3}) = (\sqrt{3})^2 + (\sqrt{3})^2 = 3 + 3 = 6$

$f(-\sqrt{3}, -\sqrt{3}) = (-\sqrt{3})^2 + (-\sqrt{3})^2 = 3 + 3 = 6$

Thus, the minimum value is

6.

2. $f(x, y) = xy$, $g(x, y) = 4x^2 + 9y^2 - 36 = 0$

$\nabla f = (y, x)$

$\nabla g = (8x, 18y)$

$y = 8\lambda x$

$x = 18\lambda y$

$4x^2 + 9y^2 = 36$

From the first two equations, $\lambda = \frac{y}{8x} = \frac{x}{18y}$, so $18y^2 = 8x^2$ and $9y^2 = 4x^2$, so $3y = \pm 2x$.

Case 1: $3y = 2x$. $x = \frac{3}{2}y$.

$4(\frac{3}{2}y)^2 + 9y^2 = 36$

$4(\frac{9}{4}y^2) + 9y^2 = 36$

$9y^2 + 9y^2 = 36$

$18y^2 = 36$

$y^2 = 2$, $y = \pm \sqrt{2}$.

If $y = \sqrt{2}$, $x = \frac{3}{2}\sqrt{2}$. If $y = -\sqrt{2}$, $x = -\frac{3}{2}\sqrt{2}$.

Case 2: $3y = -2x$. $x = -\frac{3}{2}y$.

$4(-\frac{3}{2}y)^2 + 9y^2 = 36$

$4(\frac{9}{4}y^2) + 9y^2 = 36$

$9y^2 + 9y^2 = 36$

$18y^2 = 36$

$y^2 = 2$, $y = \pm \sqrt{2}$.

If $y = \sqrt{2}$, $x = -\frac{3}{2}\sqrt{2}$. If $y = -\sqrt{2}$, $x = \frac{3}{2}\sqrt{2}$.

Possible points: $(\frac{3}{2}\sqrt{2}, \sqrt{2})$, $(-\frac{3}{2}\sqrt{2}, -\sqrt{2})$, $(-\frac{3}{2}\sqrt{2}, \sqrt{2})$, $(\frac{3}{2}\sqrt{2}, -\sqrt{2})$.

$f(\frac{3}{2}\sqrt{2}, \sqrt{2}) = \frac{3}{2}\sqrt{2} \cdot \sqrt{2} = 3$

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

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

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

The maximum value is

3.

3. $f(x, y) = 4x^2 - 4xy + y^2$, $g(x, y) = x^2 + y^2 - 1 = 0$

$\nabla f = (8x - 4y, -4x + 2y)$

$\nabla g = (2x, 2y)$

$8x - 4y = 2\lambda x$

$-4x + 2y = 2\lambda y$

$x^2 + y^2 = 1$

$4x - 2y = \lambda x$

$-2x + y = \lambda y$

$\lambda = \frac{4x - 2y}{x} = \frac{-2x + y}{y}$

$4xy - 2y^2 = -2x^2 + xy$

$3xy - 2y^2 + 2x^2 = 0$

$2x^2 + 3xy - 2y^2 = 0$

$2x^2 + 4xy - xy - 2y^2 = 0$

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

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

Case 1: $2x - y = 0$. $y = 2x$.

$x^2 + (2x)^2 = 1$

$x^2 + 4x^2 = 1$

$5x^2 = 1$

$x^2 = \frac{1}{5}$, $x = \pm \frac{1}{\sqrt{5}}$.

If $x = \frac{1}{\sqrt{5}}$, $y = \frac{2}{\sqrt{5}}$. If $x = -\frac{1}{\sqrt{5}}$, $y = -\frac{2}{\sqrt{5}}$.

Case 2: $x + 2y = 0$. $x = -2y$.

$(-2y)^2 + y^2 = 1$

$4y^2 + y^2 = 1$

$5y^2 = 1$

$y^2 = \frac{1}{5}$, $y = \pm \frac{1}{\sqrt{5}}$.

If $y = \frac{1}{\sqrt{5}}$, $x = -\frac{2}{\sqrt{5}}$. If $y = -\frac{1}{\sqrt{5}}$, $x = \frac{2}{\sqrt{5}}$.

Possible points: $(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$, $(-\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}})$, $(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}})$, $(\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$.

$f(x, y) = 4x^2 - 4xy + y^2 = (2x - y)^2$

$f(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}) = (2\frac{1}{\sqrt{5}} - \frac{2}{\sqrt{5}})^2 = 0$

$f(-\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}) = (2(-\frac{1}{\sqrt{5}}) - (-\frac{2}{\sqrt{5}}))^2 = 0$

$f(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}) = (2(-\frac{2}{\sqrt{5}}) - \frac{1}{\sqrt{5}})^2 = (-\frac{5}{\sqrt{5}})^2 = \frac{25}{5} = 5$

$f(\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}) = (2(\frac{2}{\sqrt{5}}) - (-\frac{1}{\sqrt{5}}))^2 = (\frac{5}{\sqrt{5}})^2 = \frac{25}{5} = 5$

The maximum value is

5.

4. $f(x, y) = x^2 + 4xy + y^2$, $g(x, y) = x - y - 6 = 0$

$\nabla f = (2x + 4y, 4x + 2y)$

$\nabla g = (1, -1)$

$2x + 4y = \lambda$

$4x + 2y = -\lambda$

$x - y = 6$

$2x + 4y = -(4x + 2y)$

$2x + 4y = -4x - 2y$

$6x = -6y$

$x = -y$

$x - y = 6$

$-y - y = 6$

$-2y = 6$

$y = -3$

$x = -(-3) = 3$

$f(3, -3) = (3)^2 + 4(3)(-3) + (-3)^2 = 9 - 36 + 9 = -18$

The minimum value is -

1

8.

5. $f(x, y, z) = x^2 + y^2 + z^2$, $g(x, y, z) = x + 3y - 2z - 12 = 0$

$\nabla f = (2x, 2y, 2z)$

$\nabla g = (1, 3, -2)$

$2x = \lambda$

$2y = 3\lambda$

$2z = -2\lambda$

$x + 3y - 2z = 12$

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

$y = \frac{3\lambda}{2}$

$z = -\lambda$

$\frac{\lambda}{2} + 3(\frac{3\lambda}{2}) - 2(-\lambda) = 12$

$\frac{\lambda}{2} + \frac{9\lambda}{2} + 2\lambda = 12$

$\frac{10\lambda}{2} + 2\lambda = 12$

$5\lambda + 2\lambda = 12$

$7\lambda = 12$

$\lambda = \frac{12}{7}$

$x = \frac{1}{2}(\frac{12}{7}) = \frac{6}{7}$

$y = \frac{3}{2}(\frac{12}{7}) = \frac{18}{7}$

$z = -\frac{12}{7}$

$f(x, y, z) = (\frac{6}{7})^2 + (\frac{18}{7})^2 + (-\frac{12}{7})^2 = \frac{36}{49} + \frac{324}{49} + \frac{144}{49} = \frac{504}{49} = \frac{72}{7}$

The minimum value is $\frac{72}{7}$.

6. $f(x,y,z) = 4x-2y+3z$, $g(x,y,z) = 2x^2+y^2-3z=0$

$\nabla f = (4, -2, 3)$

$\nabla g = (4x, 2y, -3)$

$4 = 4\lambda x \implies x = \frac{1}{\lambda}$

$-2 = 2\lambda y \implies y = -\frac{1}{\lambda}$

$3 = -3\lambda \implies \lambda = -1$

So $x = -1$, $y=1$, $2(-1)^2 + (1)^2 - 3z = 0$, so $2+1=3z$ then $3z = 3$, so $z = 1$.

$f(-1, 1, 1) = 4(-1) - 2(1) + 3(1) = -4 - 2 + 3 = -3$

The minimum value is -

3.

7. Let the dimensions be $x$, $y$, and $z$. The volume is $V = xyz$ and the surface area is $A = xy + 2xz + 2yz = 48$. We want to maximize $V$ subject to $A = 48$.

$\nabla V = (yz, xz, xy)$

$\nabla A = (y+2z, x+2z, 2x+2y)$

$yz = \lambda (y+2z)$

$xz = \lambda (x+2z)$

$xy = \lambda (2x+2y)$

$xy + 2xz + 2yz = 48$

From $yz = \lambda(y+2z)$ and $xz = \lambda(x+2z)$, $xyz = \lambda x(y+2z) = \lambda(xy+2xz)$ and $xyz = \lambda y(x+2z) = \lambda(xy+2yz)$. So $xy+2xz = xy+2yz$, $2xz = 2yz$ and $x=y$ (since $z>0$).

Then $x^2 = \lambda(2x+2x)$, $x^2 = 4\lambda x$, $x = 4\lambda$.

Now $x^2 + 2xz + 2xz = 48$

$x^2 + 4xz = 48$

$yz = \lambda(y+2z) \implies xz = \lambda(x+2z)$. So $xz = \lambda(x+2z)$, $xz = x/4 (x+2z)$. $4z = x+2z$ so $2z = x$.

Then $x^2 + 4x(\frac{x}{2}) = 48$

$x^2 + 2x^2 = 48$

$3x^2 = 48$

$x^2 = 16$, $x=4$.

Then $y=4$ and $z=x/2 = 4/2 = 2$.

Therefore the dimensions are 4, 4, and

2.

3. Final Answer

1. The minimum value is

6.

2. The maximum value is

3.

3. The maximum value is

5.

4. The minimum value is -

1

8.

5. The minimum value is 72/

7.

6. The minimum value is -

3.

7. The dimensions are 4, 4, and 2.