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:

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:

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.

**Answer:**

**(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:

**(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

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:

**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

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

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

Suppose that . Then exactly

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:

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 |

**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

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

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