Lagrange Multipliers
Page 793: Solutions to Problems 1, 3, 5, 9 and 11

1. Use the method of Lagrange multipliers to maximize $x^3 y^5$ subject to the constraint $x + y = 8$.

Answer: Let $f(x,y) = x^3 y^5$ and $g(x,y) = x + y - 8$. The points $(x,y)$ which will maximize $f(x,y)$ subject to the constraint $g(x,y) = 0$ will be the set of points $(x,y)$ that satisfy the equations $\nabla f(x,y) = \lambda \nabla g(x,y)$ and $g(x,y) = 0$. Writing out the system of three equations and three unknowns explicitly:

$$3 x^2 y^5 = \lambda$$ (1)

$$5 x^3 y^4 = \lambda$$ (2)

$$x + y = 8.$$ (3)

Suppose $\lambda = 0$. Then eqn.(1) would be $3 x^2 y^5 = 0$ which would imply that either $x = 0$ or $y = 0$. (They cannot both be equal to 0, because that would contradict eqn.(3).) If $x = 0$, then eqn.(3) says $y = 8$. And if $y = 0$, we would instead have from eqn.(3) that $x = 8$. Either way, we get $f(8, 0) = f(0, 8) = 0$.

Now suppose that $\lambda \neq 0$. We can set equations (1) and (2) equal to each other and simplify:

$$3 x^2 y^5 = 5 x^3 y^4$$

$$3 x^2 y^5 - 5 x^3 y^4 =$$ 0

$$x^2 y^4 (3 y - 5 x) =$$ 0 (4)

Since $\lambda \neq 0$, eqn.(1) implies that $x \neq 0$, $y \neq 0$. Therefore, for eqn.(4) to be true, it must be that $3 y - 5 x = 0$, and solving this equation for $y$: $y = \displaystyle{\frac{5}{3}} x$. Plugging this into eqn.(3) and solving for $x$ yields:

$$x + \frac{5}{3} x = 8 \Longrightarrow \frac{8}{3} x = 8 \Longrightarrow x = 3$$    and therefore $$y = 5.$$

And so we have $f(3, 5) = 3^3 \cdot 5^5 = (27)\cdot (3,125) = 84,375.$ Since clearly $f(3, 5) > f(8, 0)$    and $f(0, 8)$, $f(x,y)$ is maximized at the point $(3, 5)$.

3. Find the distance from the origin to the plane $x + 2 y + 2 z = 3,$

1.

using a geometric argument (no calculus)

2.

by reducing the problem to an unconstrained problem in two variables, and

3.

using the method of Lagrange multipliers.

Answer:

(i): From Example 7 on page 628, we know that the distance $s$ from the point $(x_0, y_0, z_0)$ to the plane whose equation is $A x + B y + C z = D$ is given by the formula:

$$s = \frac{\vert A x_0 + B y_0 + C z_0 - D\vert}{\sqrt{ A^2 + B^2 + C^2 }}.$$

And so, plugging in the appropriate numbers:

$$s = \frac{\vert(1\cdot 0) + (2\cdot 0) + (2\cdot 0) - 3\vert}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{3}{\sqrt{9}} = \frac{3}{3} = 1.$$

(ii): The distance from any point on the plane to the origin is $f(x, y, z) = \sqrt{x^2 + y^2 + z^2}$. This is the function I wish to minimize. To make Life (and taking derivatives) easier, I will instead minimize its square: $F(x,y,z) = x^2 + y^2 + z^2$. The point $(x, y, z)$ that minimizes $F(x, y, z)$ will also certainly minimize $f(x,y,z)$.

Let's solve the equation of the plane for $z$: $z = \displaystyle{\frac{1}{2}} ( 3 - x - 2 y )$, and then plug this $z$ into $F(x, y, z)$:

$$F(x, y, z) = F\left(x, y, \frac{1}{2} ( 3 - x - 2 y ) \right)$$

$$= x^2 + y^2 + {\left( \frac{1}{2} ( 3 - x - 2 y ) \right)}^2$$

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

Let that last equation equal $g(x,y)$. I now need to find the critical points of $g(x,y)$. Once I find the critical points, I plug them into $g(x,y)$ and the smallest number I get will be the square of the minimum distance. (Remember, I'm minimizing the distance-squared function.)

To find the critical points, I have to find the $(x,y)$ such that $\nabla g(x, y) = 0 {\bf i} + 0 {\bf j}$. So the equations I need to solve are

$$g_x (x, y) = -\frac{3}{2} + \frac{5}{2} x + y = 0$$

$$g_y( x, y) = -3 + x + 4 y = 0.$$

Solving the above for $x$ and $y$: $x = \displaystyle{\frac{1}{3}}$ and $y = \displaystyle{\frac{2}{3}}$, and then we get $${ g\left(\frac{1}{3}, \frac{2}{3}\right) = 1}$$, and so the distance is 1.

(iii): As in part (ii), I'll instead minimize the distance-squared: $F(x,y,z) = x^2 + y^2 + z^2$. The constraint I have is $g(x, y, z) = x + 2 y + 2 z - 3 = 0$, and this means the system of equations I need to solve is

$$2 x = \lambda$$ (5)

$$2 y = 2 \lambda$$ (6)

$$2 z = 2 \lambda$$ (7)

$$x + 2 y + 2 z = 3$$ (8)

Suppose $\lambda = 0$. Then eqns.(5-7) would imply $x = y = z = 0$. But then that would contradict eqn.(8): $0 + 0 + 0 = 3$. Therefore, we must have $\lambda \neq 0$.

Since $\lambda \neq 0$, setting eqn.(6) equal to eqn.(7) yields: $2 y = 2 z \Longrightarrow y = z$. (Note that to get this, I have to divide by $\lambda$. Since I know $\lambda \neq 0$, I can legally do so.) And using eqns.(6) and (5), I see that $y = 2 x$. Substituting all this into eqn.(8) and solving:

$$x + 2 (2 x) + 2 ( 2 x ) = 3 \Longrightarrow 9 x = 3 \Longrightarrow x = \frac{1}{3},$$

and so $${ y = \frac{2}{3} \mbox{ and } z = \frac{2}{3} }$$. And so (a big surprise here): $${F\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right) = 1}$$ and that means the minimum distance is 1.

5. Use the Lagrange multiplier method to find the greatest and least distances from the point $(2, 1, -2)$ to the sphere with the equation $x^2 + y^2 + z^2 = 1.$

Answer: The distance from $(x, y, z)$ to the point $(2, 1, -2)$ is $f(x, y, z) = \sqrt{ {(x-2)}^2 + {(y-1)}^2 + {(z+2)}^2 }$. This is the function I need to minimize and maximize subject to the constraint that $g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0$. Using the same reasoning given in the previous problem, I will instead minimize and maximize $F(x, y, z) = {(x-2)}^2 + {(y-1)}^2 + {(z+2)}^2$.

Taking the appropriate derivatives, the system of equations I need to solve is

$$2 ( x - 2 ) = 2 x \lambda x (1 - \lambda) = 2$$ which simplifies to: (9)

$$2 ( y - 1 ) = 2 y \lambda y (1 - \lambda) = 1$$ which simplifies to: (10)

$$2 ( z + 2 ) = 2 z \lambda z (1 - \lambda) = -2$$ which simplifies to: (11)

$$x^2 + y^2 + z^2 = 1 x^2 + y^2 + z^2 = 1.$$ which stays the same: (12)

Now, eqns.(9-11) all imply that $x, y, z \neq 0$. (Otherwise, we would have $0 = 2$.)

Since I know that $x, y, z \neq 0$, I can legally divide both sides of eqn.(9) by $x$, both sides of eqn.(10) by $y$ and both sides of eqn.(11) by $z$. (Before dividing, I need to first make sure that I'm not dividing by 0.) And so

$$1 - \lambda = \frac{2}{x}$$ (13)

$$1 - \lambda = \frac{1}{y}$$ (14)

$$1 - \lambda = -\frac{2}{z}$$ (15)

Setting eqns.(13) and (15) equal to each other gets:

$$\frac{2}{x} = -\frac{2}{z} \Longrightarrow z = -x$$

and from eqns.(13) and (14):

$$\frac{2}{x} = \frac{1}{y} \Longrightarrow y = \frac{x}{2}.$$

Let's plug all this into eqn.(12):

$$x^2 + {\left(\frac{x}{2}\right)}^2 + {(-x)}^2 = 1 \Longrightarrow... ... x^2 = 1 \Longrightarrow x^2 = \frac{4}{9} \Longrightarrow x = \pm\frac{2}{3}.$$

The two points we need to check are $${ \left( \frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right) }$$ and $${ \left( -\frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \right) }$$:

$$F\left( \frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right) = 4$$

$$F\left( -\frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \right) = 16.$$

Therefore, the least distance from the point $(2, 1, -2)$ to the unit sphere centered at the origin is $\sqrt{4} = 2$, and the greatest distance is $\sqrt{16} = 4$.

9. Find the maximum and minimum values of $f(x, y, z) = x y z$ on the sphere $x^2 + y^2 + z^2 = 12.$

Answer: The function we want to maximize and minimize is $f(x, y, z) = x y z$, and the constraint we have is $g(x, y, z) = x^2 + y^2 + z^2 - 12 = 0$. The system of equations we need to solve are

$$y z = 2 x \lambda$$ (16)

$$x z = 2 y \lambda$$ (17)

$$x y = 2 z \lambda$$ (18)

$$x^2 + y^2 + z^2 = 12.$$ (19)

Suppose that $\lambda = 0$. Then exactly two variables must be equal to 0. (E.g. Suppose $x = 0$ and $y, z \neq 0$. Then we'd get a contradiction from eqn.(16): $y z = 0$.) And they can't all be 0, because that would contradict eqn.(19). So no matter which variable is not equal to 0, we would have

$$f(\pm\sqrt{12}, 0, 0) = f(0, \pm\sqrt{12}, 0) = f(0, 0, \pm\sqrt{12}) = 0.$$

The $\pm\sqrt{12}$ comes from solving eqn.(19).

Now suppose that $\lambda\neq 0.$ Let me multiply both sides of eqn.(16) by $x$, both sides of eqn.(17) by $y$, and both sides of eqn.(18) by $z$, and add those three equations together:

$$3 x y z = 2 \lambda (x^2 + y^2 + z^2 )$$

$$3 x y z = 2 \lambda ( 12 )$$

$$x y z = 8 \lambda.$$ (20)

That last equation implies that $x, y, z \neq 0$ (since $\lambda \neq 0$) . Let me use eqn.(18) in eqn.(20) and solve for $z$:

$$x y z = 8 \lambda$$

$$( x y ) z = 8 \lambda$$

$$( 2 z \lambda ) z = 8 \lambda$$

$$2 z^2 = 8$$

$$z^2 = 4$$

$$z = \pm 2.$$

I can divide by $\lambda$ in that fourth line because $\lambda \neq 0$.

Similarly, I can use eqn.(17) in eqn.(20) and solve for $y$: $y = \pm 2$. And with eqn.(16) in eqn.(20), I can get $x$: $x = \pm 2$. And so there are a few possibilities. Let's make a table of the different combinations:

$x$	$y$	$z$	$f(x,y,z)$
2	2	2	8
2	2	-2	-8
2	-2	2	-8
2	-2	-2	8
-2	2	2	-8
-2	2	-2	8
-2	-2	2	8
-2	-2	-2	-8

Comparing all the numbers (e.g. $-8$, 0 and $8$), we see that the minimum value of $f(x,y,z)$ is $-8$, and the maximum value is $8$.

11. Find the maximum and minimum values of the function $f(x, y, z) = x$ over the curve of intersection of the plane $z = x + y$ and the ellipsoid $x^2 + 2 y^2 + 2 z^2 = 8$.

Answer: So we have to minimize and maximize the function $f(x, y, z) = x$ subject to the two constraints $g(x, y, z) = x + y - z = 0$ and $h(x, y, z) = x^2 + 2 y^2 + 2 z^2 - 8 = 0$. Therefore, the equations we have to solve are $\nabla f(x, y, z) = \lambda \nabla g(x, y, z) + \mu\nabla h(x,y,z)$, $g(x, y, z) = 0$, and $h(x, y, z) = 0$. That means, we solve

$$1 = \lambda + 2 x\mu$$ (21)

$$= \lambda + 4 y\mu$$ 0 (22)

$$= -\lambda + 4 z\mu$$ 0 (23)

$$x + y = z$$ (24)

$$x^2 + 2 y^2 + 2 z^2 = 8.$$ (25)

Note that it must be the case that $\mu \neq 0$. (If $\mu = 0$, then eqn.(21) would say $\lambda = 1$ and eqn.(22) would say $\lambda = 0$, and we would get a contradiction.)

Since I know I can legally divide by $\mu$, from eqns.(22) and (23) we have

$$4 y\mu = -4 z\mu \Longrightarrow y = -z$$

and using that in eqn.(24):

$$x + (-z) = z \Longrightarrow x = 2 z.$$

Let's plug all this into eqn.(25):

$${(2 z)}^2 + 2 {(-z)}^2 + 2 z^2 = 8$$

$$4 z^2 + 2 z^2 + 2 z^2 = 8$$

$$8 z^2 = 8$$

$$z^2 = 1$$

$$z = \pm 1.$$

For $z = 1$, we have $x = 2$ and $y = -1$, and $f(2,-1,1) = 2$. For $z = -1$, we have $x = -2$ and $y = 1$, and $f(-2, 1, -1) = -2$. Therefore, the minimum value is -2, and the maximum value 2.

Figure 1: Joseph-Louis Lagrange (1736-1813)