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

1. Use the method of Lagrange multipliers to maximize subject to the constraint .

Answer: Let and . The points which will maximize subject to the constraint will be the set of points that satisfy the equations and . Writing out the system of three equations and three unknowns explicitly:

 (1) (2) (3)

Suppose . Then eqn.(1) would be which would imply that either or . (They cannot both be equal to 0, because that would contradict eqn.(3).) If , then eqn.(3) says . And if , we would instead have from eqn.(3) that . Either way, we get .

Now suppose that . We can set equations (1) and (2) equal to each other and simplify:

 0 0 (4)

Since , eqn.(1) implies that , . Therefore, for eqn.(4) to be true, it must be that , and solving this equation for : . Plugging this into eqn.(3) and solving for yields:

and therefore

And so we have Since clearly     and , is maximized at the point .

3. Find the distance from the origin to the plane

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.

(i): From Example 7 on page 628, we know that the distance from the point to the plane whose equation is is given by the formula:

And so, plugging in the appropriate numbers:

(ii): The distance from any point on the plane to the origin is . This is the function I wish to minimize. To make Life (and taking derivatives) easier, I will instead minimize its square: . The point that minimizes will also certainly minimize .

Let's solve the equation of the plane for : , and then plug this into :

Let that last equation equal . I now need to find the critical points of . Once I find the critical points, I plug them into 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 such that . So the equations I need to solve are

Solving the above for and : and , and then we get , and so the distance is 1.

(iii): As in part (ii), I'll instead minimize the distance-squared: . The constraint I have is , and this means the system of equations I need to solve is

 (5) (6) (7) (8)

Suppose . Then eqns.(5-7) would imply . But then that would contradict eqn.(8): . Therefore, we must have .

Since , setting eqn.(6) equal to eqn.(7) yields: . (Note that to get this, I have to divide by . Since I know , I can legally do so.) And using eqns.(6) and (5), I see that . Substituting all this into eqn.(8) and solving:

and so . And so (a big surprise here): and that means the minimum distance is 1.

5. Use the Lagrange multiplier method to find the greatest and least distances from the point to the sphere with the equation

Answer: The distance from to the point is . This is the function I need to minimize and maximize subject to the constraint that . Using the same reasoning given in the previous problem, I will instead minimize and maximize .

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

 which simplifies to: (9) which simplifies to: (10) which simplifies to: (11) which stays the same: (12)

Now, eqns.(9-11) all imply that . (Otherwise, we would have .)

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

 (13) (14) (15)

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

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

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

The two points we need to check are and :

Therefore, the least distance from the point to the unit sphere centered at the origin is , and the greatest distance is .

9. Find the maximum and minimum values of on the sphere

Answer: The function we want to maximize and minimize is , and the constraint we have is . The system of equations we need to solve are

 (16) (17) (18) (19)

Suppose that . Then exactly two variables must be equal to 0. (E.g. Suppose and . Then we'd get a contradiction from eqn.(16): .) 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

The comes from solving eqn.(19).

Now suppose that Let me multiply both sides of eqn.(16) by , both sides of eqn.(17) by , and both sides of eqn.(18) by , and add those three equations together:

 (20)

That last equation implies that (since ) . Let me use eqn.(18) in eqn.(20) and solve for :

I can divide by in that fourth line because .

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

 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. , 0 and ), we see that the minimum value of is , and the maximum value is .

11. Find the maximum and minimum values of the function over the curve of intersection of the plane and the ellipsoid .

Answer: So we have to minimize and maximize the function subject to the two constraints and . Therefore, the equations we have to solve are , , and . That means, we solve

 (21) 0 (22) 0 (23) (24) (25)

Note that it must be the case that . (If , then eqn.(21) would say and eqn.(22) would say , and we would get a contradiction.)

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

and using that in eqn.(24):

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

For , we have and , and . For , we have and , and . Therefore, the minimum value is -2, and the maximum value 2.