EXAMPLE 1:

Find the point P (x, y, z) closest to the origin on the plane x + y - z = 2.

SOLUTION:

The problem is asking us to find the minimum value of the function |OP|, which is the length of the vector from the origin to P subject to the constraint x + y - z = 2.

A fact that we should keep in mind is that |OP| will have a minimum value wherever the function f (x, y, z) = x ² + y ² + z ² has a minimum value. So the problem turns into find the minimum value of f (x, y, z) = x ² + y ² + z ² subject to the constraint x + y - z = 2.

If we regard x and y as independent variables, then z = x + y - 2, and a new function h (x, y) = f (x, y, x + y - 2) = x ² + y ² + (x + y - 2) ².

 Well, that was not to bad, so let us do another example.

 EXAMPLE 2:

Find the points closest to the origin on the hyperbolic cylinder x ² - z ² - 1 = 0. 

SOLUTION A:

Again f (x, y, z) = x ² + y ² + z ², and we want to minimize f subject to the constraint x ² - z ² - 1 = 0. If we regard x and y as the independent variables, then z ² = x ² - 1 and

The critical point is (0, 0), but when I plug it into z ² = x ² - 1, I get z ² = -1 and that cannot happen. In fact, (0, 0) is not on the cylinder. We did not choose our independent variables wisely. If we look at the surface x ² - z ² = 1, we should have noticed that the domain for this surface is in the xz-plane. Therefore, let y and z be the independent variables, and x ² = z ² + 1.

k (y, z) = z ² + 1 + y ² + z ² = 2z ² + y ² + 1

The closest points are (± 1, 0, 0) and the distance is 1.

As you can see, this method for finding the max or min values subject to a constraint can have many pitfalls. There has to be a better way, and there is.

SOLUTION B:

Another way to solve this problem is to imagine a small sphere centered at the origin. This sphere is expanding like a soap bubble until it touches the cylinder x ² - z ² - 1 = 0. Now f (x, y, z) = x ² + y ² + z ² - a ² (the sphere) and g (x, y, z) = x ² - z ² - 1 (the cylinder).

FACT: Ñ f will be parallel to Ñ g at the points where the sphere touches the cylinder. Therefore Ñ f = l Ñ g where l is a scalar.

Now, the question is this: for what values of l will the point (x, y, z) whose coordinates satisfy the above equations will also lie on the surface x ² - z ² - 1 = 0? Well, if x = 0, then -z ² = 1 which cannot happen. When x ¹ 0, then 2 = 2l ® l = 1. When l = 1 ® 2z = -2z, and for this to be true, z has to equal zero. y = 0 (2y = 0) and z = 0, so the point closest to the origin has the form (x, 0, 0). Since z = 0 ® x ² -1 = 0 ® x ² = 1 and x = ± 1.

Therefore the points closest to the origin are (± 1, 0, 0).

This way of solving this problem is called the method of Lagrange Multipliers. This method states that the extreme values of a function f (x, y, z) whose variables are subject to the constraint g (x, y, z) = 0 are to be found on the surface g = 0 at the points where Ñ f = l Ñ g for some scalar l . l is called the Lagrange multiplier.

What is the theory that makes this method work? If is based on a theorem and a corollary.

THEOREM:

Suppose that f (x, y, z) is differentiable in a region whose interior contains a smooth curve C: r = g (t) i + h (t) j + k (t) k. If P₀ is a point on C where f has a local maximum or minimum relative to its value on C, then Ñ f is orthogonal to C at P₀.

COROLLARY:

At points on a smooth curve r = g (t) i + h (t) j where a differentiable function f (x, y) takes on its local maxima and minima relative to its value on the curve, Ñ f · v = 0.

Both of these facts are important to the method of Lagrange multipliers. How they are important to this method is on how we can relate the two gradient vectors. Suppose that f (x, y, z) and g (x, y, z) are differentiable and that P₀ is a point on the surface g (x, y, z) = 0 where f has a local maximum or minimum relative to its other values on the surface. Then f takes on a local max or min at P₀ relative to its values on every differentiable curve through P₀ on the surface f (x, y, z) = 0. Therefore, Ñ f is orthogonal to the velocity vector of every differentiable curve through P₀, and so is Ñ g. Therefore, at P₀, Ñ f is a scalar multiple l of Ñ g.

 Suppose that f (x, y, z) and g (x, y, z) are differentiable. To find the local maximum and minimum values of f subject to the constraint g (x, y, z) = 0, find the values of x, y, z and l that simultaneously satisfy the equations Ñ f = l Ñ g and g (x, y, z) = 0. For functions of two independent variables, Ñ f = l Ñ g and g (x, y) = 0.

 EXAMPLE 3:

 Find the extreme values of f (x, y, z) xy subject to the constraint g (x, y) = x ² + y ² - 10 = 0.

SOLUTION:

CASE 1: If x = 0, then y = 0, but (0, 0) is not on the circle x ² + y ² = 10. Therefore, x ¹ 0.

CASE 2: x ¹ 0 and l = ± 1/2

EXAMPLE 4:

Find the points on the curve xy ² = 54 nearest the origin.

SOLUTION:

Since we want the points nearest the origin, let f be the square of the distance from the origin to a point on the curve.

CASE 1: y = 0 ® x = 0, but (0, 0) does not satisfy the constraint. Therefore, y ¹ 0.

CASE 2: y ¹ 0 and y ² = 2/l ².

So the points nearest the origin are

EXAMPLE 5:

Find the maximum and minimum values of x ² + y ² subject to the constraint x ² - 2x + y ² - 4y = 0.

SOLUTION:

5l ² - 10l ² + 10l = 0 ® -5l ² + 10l = 0 ® -5l (l - 2) = 0 ® l = 0 or l = 2

CASE 1: l = 0 ® x = 0 and y = 0. g (0, 0) = 0 is true. f (0, 0) = 0 is a minimum.

CASE 2: l = 2 ® x = -2/(1 - 2) = 2 and y = -4/(1 - 2) = 4. g (2, 4) = 4 - 4 + 16 - 16 = 0 is true. f (2, 4) = 4 + 16 = 20 is the maximum.

 EXAMPLE 6:

Find the point on the plane x + 2y + 3z = 13 closest to the point (1, 1, 1). 

 SOLUTION:

Let P (x, y, z) be a point on the plane. Define f to be the square of the distance from the point (1, 1, 1) and P. Therefore, f = (x - 1) ² + (y - 1) ² + (z - 1) ². Let g = x + 2y + 3z - 13.

(3/2, 2, 5/2) is the point on the plane closest to the point (1, 1, 1).

 EXAMPLE 7:

Find the point on the surface z = xy + 1 nearest the origin. 

 SOLUTION:

Let P (x, y, z) be a point on the surface z = xy + 1 and f be the square of the distance from P to the origin. Then f = x ² + y ² + z ² and g = xy - z + 1.

CASE 1: y = 0 ® x = 0 ® z = 0. Therefore, the point is (0, 0, 1)

CASE 2: l = 2 ® 2z = -2 ® z = -1 and x = y.

z = xy + 1 ® -1 = x ² + 1 ® x ² = -2. This cannot happen, therefore no solution.

CASE 3: l = -2 ® 2z = 2 ® z = 1 and x = -y.

z = xy + 1 ® 1 = -y ² + 1 ® y ² = 0 ® y = 0 and x = 0.

The point is the same point that we found in case 1.

The point (0, 0, 1) is the point on the surface z = xy + 1 closest to the origin.

EXAMPLE 8:

Find the dimensions of a closed rectangular box with maximum volume that can be inscribed in the unit sphere.

SOLUTION:

Assume that the sphere and the box are centered at the origin. The length of the box is 2y. The width of the box is 2x, and the height is 2z. Therefore, the volume will be V = 8xyz. Let g = x ² + y ² + z ² -1, which is the unit sphere.

 The dimensions of the box are

 EXAMPLE 9:

 Minimize the function f (x, y, z) = x ² + y ² + z ² subject to the constraints x + 2y + 3z = 6 and x + 3y + 9z = 9.

SOLUTION:

Now plug these equations into g₁ and g₂.

Now I am going to solve these two equations simultaneously.