f(x,y) = 3x^2 + 3yx^2 + y^3 -15y.
Step 1. Take the gradient:
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:
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
also a saddle.
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.
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
To find the best dimensions, we need Lagrange's formula:
Grad f == lambda Grad g, i.e.,
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.