Math 212 Spring 2008: Solutions: HW #6
Instructor: S. Cautis
1. section 3.3, #2
∂f /∂x = 2x − y and ∂f /∂y = 2y − x. These are both zero when x = y = 0
so the only critical point is (0, 0).
Now ∂ 2 f /∂x2 = 2, ∂ 2 f /∂x∂y = −1 while ∂ 2 f /∂y 2 = 2. So the Hessian is
2 −1
−1 2
This is positive difinite at (0, 0) since the upper left entry as well as the
determinant are positive. So (0, 0) is a local minimum.
2. section 3.3, #4
∂f /∂x = 2x + 3y and ∂f /∂y = 2y + 3x. These are both zero when
x = y = 0 so the only critical point is (0, 0).
Now ∂ 2 f /∂x2 = 2, ∂ 2 f /∂x∂y = 3 while ∂ 2 f /∂y 2 = 2. So the Hessian is
2 3
3 2
The determinant of this matrix is −5 < 0. This means that the point
(0, 0) is a saddle point.
3. section 3.3, #10
∂f /∂x = sin y and ∂f /∂y = 1 + x cos y. The first is zero when y = πk
where k is any integer. Now cos(πk) equals 1 if k is even and −1 is k is odd.
So we get an infininte number of critical points, namely (x, y) = (−1, πk)
if k is even and (x, y) = (1, πk) when k is odd.
Now ∂ 2 f /∂x2 = 0, ∂ 2 f /∂x∂y = cos y while ∂ 2 f /∂y 2 = −x sin y. So the
Hessian is
0
cos y
cos y −x sin y
The determinant is − cos2 y so when evaluated at any of the critical points
this is negative. Therefore all the critical points are saddle points.
4. section 3.3, #22
A point in the plane 2x − y + 2z = 20 looks like
p (x, 2x + 2z − 20, z) where
we substituted for y. We want to minimize x2 + y 2 + z 2 . This is the
same as minimizing x2 + y 2 + z 2 which should be computationally easier
to do. Substituting for y we get the function
g(x, z) = x2 + (2x + 2z − 20)2 + z 2
where x, z can be any real numbers.
Let’s find the critical points. ∂g/∂x = 2x+2(2x+2z−20)·2 = 10x+8z−80
while ∂g/∂z = 2(2x + 2z − 20) · 2 + 2z = 8x + 10z − 80. Solving
10x + 8z − 80 = 8x + 10z − 80 = 0
we get x = 5, y = 5 as the only critical point.
Now the Hessian can easily be computed to equal
10 8
8 10
which is positive definite. So (5, 5) is a minimum for g(x, z) and hence the
closest point to the origin is
(x, y, z) = (5, 0, 5)
5. section 3.3, #30
The only critical point inside is (x, y) = (0, 0). On the boundary of the
disk x2 + y 2 = 1 so f takes the value 1 everywhere on the boundary. So
the minimum value is 0 and is achieved at (0, 0) while the maximum value
is 1 which is achieved everywhere on the boundary of the unit disk.
6. section 3.3, #32
We can parametrize the cylinder x2 + y 2 = 1 as (cos t, sin t, r) where 0 ≤
t ≤ 2π and r is any real number. The the additional equation imposes the
condition
cos2 t − cos t sin t + sin2 t − r2 = 1
which simplifies to r2 = − cos t sin t. We want to minimize
x2 + y 2 + z 2 = cos2 t + sin2 t + r2 = 1 + r2
subject to the condition that r2 = − cos t sin t. In other words, we want
to minimize g(t) = 1 − cos t sin t where 0 ≤ t ≤ 2π.
But we need to be a bit careful here. Since r2 = − cos t sin t we need
that − cos t sin t ≥ 0. This means precisely one of cos t, sin t should be
negative and the other positive (or equal to zero). This occurs exactly
when ≤ π/2 ≤ t ≤ π and 3π/2 ≤ t ≤ 2π.
Now g ′ (t) = sin2 t − cos2 t which is zero when sin2 t = cos2 t. This is the
case when tan2 t = ±1 which happens when t = π/4, 3π/4, 5π/4, 7π/4.
So these are the critical points. Only the second and the fourth critical
points lie in the allowed intervals.
So we just need to check the values of g at the two critical points and at
the endpoints 0, π/2, π, 3π/2. We have g(3π/4) = 1 − (−1/2) = 3/2 and
g(7π/4) = 1 − (−1/2) = 3/2 while g(0) = g(π/2) = g(π) = g(3π/2) = 0.
So the minimum occurs when t = 0 which corresponds to the point (1, 0, 0).
7. section 3.3, #34
∂f /∂x = y and ∂f /∂y = x so the only critical point is (0, 0) where
f (0, 0) = 0. Now there are four sides to the boundary of R. The first side
is (t, 1) where −1 ≤ t ≤ 1. The function f restricted to this side equals
f (t, 1) = t which is maximized when t = 1 and minimized when t = −1
with values 1 and −1 respectively.
Similarly, on the side (t, −1) where −1 ≤ t ≤ 1 the function is −t which
has maximum 1 and minimum −1. The third and fourth sides have similar
maxima 1 and minima −1.
So the global maximum is 1 (achieved at the top right and bottom left
corners of R) and the global minimum is −1 (achieved at the top left and
bottom right corners of R).
8. section 3.3, #36
Suppose (x0 , y0 ) is a critical point of u inside D.
2
If ∂∂xu2 (x0 , y0 ) ≥ 0 then the Hessian is not negative definite and so (x0 , y0 )
cannot be a maximum.
2
Similarly, if ∂∂yu2 (x0 , y0 ) ≥ 0 then the Hessian is not negative definite and
so (x0 , y0 ) cannot be a maximum.
2
2
Finally if we had ∂∂xu2 (x0 , y0 ) < 0 and ∂∂yu2 (x0 , y0 ) < 0 then their sum
would be negative which would contradict that u is strictly subharmonic.