Calculus 3-Fall 2012
Quiz 6, October 11, 2012
Solutions
INSTRUCTIONS: The usual instructions apply. Above all:
SHOW ALL YOUR WORK!!!
1. Find all critical points and classify them as local maximum, minimum or saddle point, of the function
f (x, y) = y 3 + 3x2 y − 6x2 − 6y 2 + 2.
Solution.
Finding the critical points. We differentiate with respect to x and to y and set the derivatives to 0:
fx
fy
= 6xy − 12x = 0
= 3y 2 + 3x2 − 12y = 0.
The first of these equations can be written in the form 6x(y − 2x) = 0; it has two solutions x = 0, y = 2. Assume first
x = 0; then fx = 0. What happens with fy ?; that is, how does the second equation look? With x = 0, the second
equation becomes 3y 2 − 12y = 0, equivalently 3y(y − 4) = 0. There are two solutions, y = 0 and y = 4. We found two
critical points: (0, 0) and (0, 4). These are the only ones with x = 0. Assume now x ̸= 0 and y = 2 (so, again, fx = 0).
With this value of x the second equation becomes 12 + 3x2 − 24 = 0 or x2 = 4, thus x = ±2. The critical points are:
(0, 0), (0, 4), (−2, 2), and (2, 2).
Classification. The Hessian determinant is
6y − 12
6x
D(x) = 6x 6y − 12
= (6y − 12)2 − 36x2 .
Then D(0, 0) = 144 > 04, fxx = −12 < 0; (0, 0) is a local maximum;
D(0, 4) = 144 > 0, fxx (0, 4) = 12 > 0; (0, 4) is a local minimum;
D(−2, 2) = D(2, 2) = −144 < 0; (±2, 2) are saddle points.
2. Find the absolute maximum and minimum VALUES of the following function f on the given set D.
f (x, y) = 2x3 + y 4 ,
D = {(x, y) : x2 + y 2 ≤ 1}.
Incidentally, value means value. The maximum (or minimum) value should be a number,
Hint: the boundary of D can be described also as {(x, y) : x = cos θ, y = sin θ, 0 ≤ θ ≤ π}.
Solution. Its easy to see that the only critical point of f (x, y) is (0, 0), so a potential maximum/minimum value
could happen there. Next we look at what happens on the boundary. The boundary is perhaps best described as the
set of points x = cos θ, y = sin θ, 0 ≤ θ ≤ 2π. The boundary values coincide thus with the values of
g(θ) = 2 cos3 θ + sin4 θ.
We apply Calculus 1 to find where g is maximum/minimum in the interval [0, 2π]. Setting g ′ (θ) = 0 gives
−6 cos2 θ sin θ + 4 sin3 θ cos θ = 0,
i.e.2 cos θ sin θ(3 cos θ − 2 sin2 θ) = 0.
so g ′ (θ) = 0 if and only if cos θ = 0 or sin θ = 0 or 3 cos θ − 2 sin2 θ = 0. The first two possibilities work out to
θ = 0, π/2, π, 3π/2, or 2π.
The points 0, 2π are the endpoints of the interval, and would have been looked at anyway. To solve 3 cos θ − 2 sin2 θ = 0,
replace sin2 θ by 1 − cos2 θ to get 2 cos2 θ + 3 cos θ − 2 = 0. This is a quadratic equation in cos θ with the solution
√
−3 ± 25
1
cos θ =
= −2, .
4
2
The only solution that makes sense is cos θ = 12 , which happens for θ = π/3, 5π/3.
Looking at all boundary values where there could be a maximum/minimum:
g(0) = 2,
g(π/3) =
13
,
16
g(π/2) = 1,
g(π) = −2,
g(3π/2) = 1,
g(5π/3) =
13
,
16
g(2π) = 2.
The maximum value on the boundary is 2, the minimum −2. Evaluating at our only critical point, we have f (0, 0) = 0,
so that doesn’t change anything. The maximum value is 2, the minimum −2.
Alternative Problem (In lieu of # 2) Find the absolute maximum and minimum values of f (x, y) = x2 + y 2 + x2 y + 4
on the set D = {(x, y) : |x| ≤ 1, |y| ≤ 1}.
Solution.
Setting fx = 0, fy = 0 results in the equations 2x + 2xy = 0 and 2y + x2 = 0. The only solutions are (0, 0),
√
(± 2, −1). Of these, only (0, 0) is in the region, so it is the only one that matters. We notice f (0, 0) = 4.
Next we have to analyze what happens on the boundary. The boundary can be divided into 4 segments, namely:
I. x = 1, −1 ≤ y ≤ 1.
II. y = 1, −1 ≤ x ≤ 1.
III. x = −1, −1 ≤ y ≤ 1.
IV. y = −1, −1 ≤ x ≤ 1.
We look at the behavior on each segment.
I. We consider g(y) = f (1, y) = 5 + y + y 2 , −1 ≤ y ≤ 1. We have g ′ (y) = 1 + 2y, so g ′ (y) = 0 for y = −1/2. Now g(−1) = 5,
g(−1/2) = 4.75, g(1) = 7. The maximum on this segment is 7, the minimum 4.5.
II. We consider g(x) = f (x, 1) = 5 + 2x2 , −1 ≤ x ≤ 1. Then g ′ (x) = 4x is 0 for x = 0. Since g(−1) = g(1) = 7, g(0) = 5.
On this segment the maximum is 7, the minimum 5.
III. Now we look at g(y) = f (−1, y) = 5 + y + y 2 , −1 ≤ y ≤ 1. This is exactly the same function as in I, so the maximum
here is again 7, minimum 4.5.
IV. The function to consider is g(x) = f (x, −1) = 5, −1 ≤ x ≤ 1. The function is constantly equal to 5.
Looking at all potential maxima and minima, we see that the absolute maximum is 7, the absolute minimum is 4.
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