Mathematics 2507 Test number 1 solutions Thursday, 6 October 1994

1 (5 points). Find and classify all the critical points of the function

f(x,y) = 3x^2 + 3yx^2 + y^3 -15y.

Grad f = (6 x + 6 x y) i + (3 x^2 + 3 y^2 - 15) j

This is always defined, so the question is whether the gradient is 0:

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In[1]:= Solve[{6 x + 6 x y == 0, 3 x^2 + 3 y^2 - 15 == 0}, {x,y}]

Out[1]= {{x -> -2, y -> -1}, {x -> 2, y -> -1}, {y -> -Sqrt[5], x -> 0},

>    {y -> Sqrt[5], x -> 0}}

To classify them, we evaluate the discriminant:

f_xx f_yy  - (f_xy)^2 == (6 + 6 y)(6 y) - (6 x) (6 x)

For the first critical point, this is (-6)(-6) - (-12) (-12) < 0,
so the first critical point is a saddle.

For the second critical point, the discriminant is the same, so it is

For the third critical point, the discriminant is 5 - 0 > 0, so it is
an extremum.  Since f_xx < 0, this is a local maximum.  It cannot be
a global maximum, because for very large y (positive or negative),
the most important term in f is y^3, and it could be either positive or
negative.

For the final critical point, the discriminant is likewise positive, but
f_xx > 0, so it is a (local) minimum.

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2 (5 points). A cylindrical can of volume 20 cm^3 is to be constructed. Material for the top and bottom costs \$0.01/cm^2, and for the side \$0.005/cm^2. What dimensions R=radius and H=height should the can have to minimize the cost?

The objective function, is the cost:

f(R,H) == 2 * pi * R^2 * .01 + (2 * pi * R) H * .005

The constraint is the volume:

g(R,H) = pi * R^2 * H == 20

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To find the best dimensions, we need Lagrange's formula:

R component:

.04 * pi * R  + .01 * pi * H == lambda * 2 * pi * R * H

H component:

.01 * pi * R  == lambda * pi * R^2

To solve, let's multiply these equations by 100/pi, to get:

In[2]:= Solve[{4 R + H == 200 lambda R H, R == 100 lambda R^2, \
Pi R^2 H == 20}, {R,H, lambda}]

Out[2] =

1/3
(25 Pi)             5  1/3        5  1/3
>    {lambda -> ----------, H -> 4 (--)   , R -> (--)   }}
500              Pi            Pi

The cost of the can is:

In[3]:= .02 * Pi * R^2 + .01 * Pi *R * H   /. %

1/3
Out[3]= 0.06 (25 Pi)

3.  (5 points - Dang if this doesn't look a lot like homework problem
21!)  Find the minimum value of x^3 + y^3 + z^3
for (x,y,z) on the intersection of the planes x + y + z = 2 and x - y + z =
3.

The place(s) where the minimum occurs is (are) where LAgrange's condition
is satisfied.

With two constraints, we have:

Grad(x^3 + y^3 + z^3) = lambda Grad(x + y + z) + mu Grad(x - y + z)

The three components of this read:

3 x^2 == lambda + mu
3 y^2 == lambda - mu
3 z^2 == lambda + mu

There are 5 unknowns, so we need two more equations, which are the
constraint equations.  Comparing the equations given above, we
can see that x^2 == z^2, so z = x or -x.  Suppose first that z == x.
From the constraint equations,  we then learn that y == 2 - 2 x and
-y == 3 - 2 x.  Adding these together gives 0 == 5 - 4 x, so one
constrained critical point is
(x,y,z) == (5/4, -1/2, 5/4)
The other possibility is that z == -x.  The difficulty  with this possibility is that it
converts the constraint equations into y == 2 and -y == 3, which is impossible.
Thus there is only one critical point.

A good student will wonder here if the critical point is actually a
minimum.  It is, as can be seen by substituting the constraint
equations into the objective, which becomes a quadratic representing a
parabola going to +infinity.  (We would probably give almost total
credit without a carefgul proof of this point.)

The value of the minimum is 121/32.

4.  (5 points)  NOTE:  The point of this problem is to see if you
understand Newton's method.  The exact solution of the system is  (x,y) = (3,2)
or (2,3).  Now that we all know the exact answer, you will get 0 points
for writing it down!

Suppose

f(x,y) := x + y - 5 = 0 and
g(x,y) := (x - y)^2 - 1 = 0.

Take as your initial guess (x_0,y_0) = (1,-1) and use Newton's method to produce a
better guess (x_1,y_1)

The explicit* formula for the improved guess (x_{n+1},y_{n+1}) given  (x_n,y_n) is:

x_{n+1} = _________________________________________________

y_{n+1} = _________________________________________________

With (x_0,y_0) = (1,1),

x_1 = _________________________________________________

y_1 = _________________________________________________

This problem is a "plug and chug," and the answer can be looked up in the
textbook.

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