**1** (10 points) Find the point of the ellipse
defined by

5 x^^{2} + 5 y^^{2} - 6 xy - 2 x - 2 y + 3/4 = 0

which is closest to the origin.

SOLUTION

The list of "candidate" points is determined by looking for constrained critical

points with Lagrange's formula. The objective function is the distance or,

more conveniently, the square of the distance from the origin,

f(x,y) == x^2 + y^2.

Lagrange's formula states that the constrained critical points are solutions of

Grad[f] == lambda Grad[g],

where g is the function defining the ellipse. So here:

2 x **i** + 2 y **j** == lambda ( (10 x - 6 y - 2) **i** + (10 y
- 6 x - 2) **j **).

This is equivalent to two scalar equations,

2 x == lambda (10 x - 6 y - 2)

2 y == lambda (10 y - 6 x - 2)

The next step is to eliminate lambda. As in some of the homework problems, the

efficient way to do this is to cross multiply by convenient factors. In this case,

multiply the first equation by y/lambda and the second by x/lambda, so that the

left sides are equal. (It is legitimate to divide by lambda since it is not 0 -

if lambda == 0, then x = y = 0, but this is inconsistent with the constraint.)

When we equate the right sides, we get:

10 x y - 6 y^2 - 2 y == 10 x y - 6 x^2 - 2 x.

The 10xy terms drop out and the equation simplifies to:

0 == 3 (x^2 - y^2) + (x-y) == (3 (x+y) + 1) (x-y)

There are two possibilities.

A. x - y == 0. If this is true, then y == x, and the constraint equation becomes

0 == 4 x^2 - 4 x + 3/4

which is a quadratic with roots x = 1/4 and 3/4. The "candidate" points are

(1/4, 1/4) and (3/4,3/4).

Obviously the closest point is (1/4, 1/4), and the distance from the origin is 1/(2 Sqrt[2]),

which is numerically about .353.

B. 3 (x + y) + 1 == 0, so y == - x - 1/3. Substitute into the constraint to

get a quadratic:

16 x^2 + (16/3) x + 71/36 == 0,

but this equation has no real roots, since b^2 - 4 a c < 0. Possibility B produces

no additional candidates.

**2**. (10 points - dang if this isn't like a homework problem from p.
902!)

Find the critical points of the function

f(x,y) = 3 y^^{3} + x^^{2} - 3 xy - 1

and determine their nature. If they are extrema, state whether local or global.

SOLUTION. Take the gradient and set both components to 0:

Grad[f] == (2 x - 3 y) **i** + (9 y^2 - 3 x) **j**

== 0 when y == 2 x /3 and thus 4 x^2 - 3 x == 0. The solutions of this are

(x,y) == (0,0) and (x,y) == (3/4,1/2)

We can use the second derivative test (p. 898 in Grossman) to find that

(0,0) is a saddle and (3/4,1/2) is a local extremum.

Since f_xx > 0 at (3/4,1/2), it is a local minimum. It is not a global minimum,

since, for example,

f(0, -1000) = - 3,000,000,001 < f(3/4,1/2) = - 9/16

**3**. (10 points) Sir Isaac Newton is snowboarding on a snowy island,
which is

shaped like the graph of

z = f(x,y) := 1 - (x^^{2} + 4 y^^{2})/100

Distances are measured in k.m. and the sea is at height z=0. He is

currently at position

(x,y) = (6,1)

a) He drops his apple. Which direction does it roll? One of the most basic

ideas we have discussed is that the gradient points uphill. Apples roll downhill,

so that direction would be opposite to the gradient. The negative of the gradient

of f at position (6,1) is:

**v**apple = (3/25) **i** + (2/25) **j **

b) Find the formula for the (tangent) plane which contains his snowboard.

You were repeatedly reminded that it was wise to know the formula for a

normal vector to a surface z == f[x,y],

**N** = - f_x **i** - f_y **j** + **k** = (3/25) **i**
+ (2/25) **j** + **k**

for this function. The plane is oriented perpendicularly to this vector

and passes through the 3D point (6,1, 0.6). (You plug (x,y) == (6,1)

into f to find the height 0.6 .) The formula for the tangent plane is

(3/25) (x - 6) + (2/25) (y - 1) + (z - 0.6) == 0.

c) What is the nearest point of the sea which is also in this tangent plane?

There are several ways to answer this, but the easiest is to remember that the

direction from Newton to the nearest point of this plane in the sea is the negative

of the gradient. We thus have two equations:

i) The plane intersects the sea when z == 0, so

(3/25) (x - 6) + (2/25) (y - 1) == 0.6

ii) The line through Newton's position (6,1) in the direction of the gradient has slope

2/3 (y will increase by 2/25 when x increases by 3/25). Its formula is:

y == 2 x/3 - 3 (use point-slope and solve for y)

Now solve for the intersection:

`In[1]:= Solve[{y == 2 x/3 - 3, \`

In[1]:= Solve[{y == 2 x/3 - 3, \ (12(x - 6) + 8 (y - 1))/100 == 3/5}, {x,y}] 123 43 Out[1]= {{x -> ---, y -> --}} 13 13 In[2]:= N[%] Out[2]= {{x -> 9.46154, y -> 3.30769}}

Extra credit: From the point found in part 3, which way should Newton swim to

get back to the island?

There are two acceptable answers to the extra credit question.

1. Newton would swim "up gradient" (since the land is at a height below z level).

The gradient at the point found in part c) is

- (246/13)**i** - (344/13)**j** ,

or, numerically:

-18.9231**i** - 26.4615**j** ,

so he would swim in this direction.

2. If Newton consulted Lagrange, before beginning to swim he would solve

a more difficult problem before starting to swim: He would find the closest

point of the shore and swim towards it. This would mean setting up an

optimization problem to minimize the objective (x - 123/13)^2 + (y - 43/13)^2

subject to the constraint 1 - (x^2 + 4 y^2)/100 == 0. Mathematica can

find an exact solution to this with the command

`In[3]:= Solve[{2 (x - 123/13) == lambda (- x/50), \`

` 2 (y - 43/13) == lambda (- 2 y/25), \`

` 1 - (x^2 + 4 y^2)/100 == 0}, {x,y,lambda}]`

but it is long and involved. Extra credit will be given for setting the problem up

properly this way.

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