Quiz 2 for MATH 105 SECTION 205
January 21, 2015
Family Name
Given Name
Student Number
1. (a) (1 point) Let f (x, y) = y sin(xy), compute fyx (x, y).
(a)
Answer.
2y cos(xy) − xy 2 sin(xy)
First Approach:
fy (x, y) = sin(xy) + y cos(xy) · x
By the product rule and the chain rule
= sin(xy) + xy cos(xy)
fyx (x, y) = cos(xy) · y + y cos(xy) − xy sin(xy) · y
By the chain rule, the product rule and the chain rule
2
= 2y cos(xy) − xy sin(xy)
Second Approach:
fx (x, y) = y cos(xy) · y
By the chain rule
2
= y cos(xy)
fyx (x, y) = fxy (x, y)
By Clairaut’s theorem
= 2y cos(xy) − y 2 sin(xy) · x
By the product rule and the chain rule
2
= 2y cos(xy) − xy sin(xy)
(b) (1 point) Can you find a function h(x, y) such that hxx = y 2 cos(xy 2 ) and hxy = xy 2 cos(xy 2 )? (Do not
need to find h(x, y), just put ‘Yes’ or ‘No’)
(b)
Answer.
No
If such a function h(x, y) exists, by the Clairaut’s theorem, we must have
hxxy = hxyx .
Since hxx = y 2 cos(xy 2 ), then
hxxy = 2y cos(xy 2 ) − y 2 sin(xy 2 ) · 2xy
2
3
By the product rule and the chain rule
2
= 2y cos(xy ) − 2xy sin(xy ).
Since hxy = xy 2 cos(xy 2 ), then
hxyx = y 2 cos(xy 2 ) − xy 2 sin(xy 2 ) · y 2
2
2
4
By the product rule and the chain rule
2
= y cos(xy ) − xy sin(xy ).
So we get hxxy 6= hxyx , contradiction.
2. (a) (0.5 points) Let g(x, y) = 2x2 + y 4 + 1, find a local minimum point of g(x, y).
(a)
Answer.
(0, 0)
It’s easy to see that
g(x, y) = 2x2 + y 4 + 1 ≥ 0 + 0 + 1 = 1 = g(0, 0).
So (0, 0) is a local minimum point of g(x, y). In fact, this is the only one minimum point of g(x, y) in R2 .
(b) (1 point) Let R = {(x, y) : x2 + 2y 2 ≤ 9} and f (x, y) = x2 + 4y 2 , find the absolute maximum value of
f (x, y) on R.
18
(b)
Answer.
For all (x, y) in R, we know that
x2 + 2y 2 ≤ 9,
and
− 3 ≤ x ≤ 3.
Notice that
f (x, y) = x2 + 4y 2 = 2(x2 + 2y 2 ) − x2 .
So for the first term 2(x2 + 2y 2 ), it has the largest possible value 2 · 9 = 18 which can be achieved for
every point on the boundary of R. For the second term −x2 , it has the largest possible value 0 which can
be achieved if x = 0. In summary, f (x, y) can achieve its absolute maximum on the boundary of R with
3
the condition x = 0, that is, y = ± √ . So the absolute maximum value of f (x, y) on R is 18 + 0 = 18.
2
3. Let R = {(x, y) : x2 + y 2 ≤ 4} and f (x, y) = 2x2 + 2y 2 + 1, then
(a) (1 point) Find the absolute maximum value of f (x, y) on R.
9
(a)
Answer.
For all (x, y) in R, we know that
x2 + y 2 ≤ 4.
Notice that
f (x, y) = 2x2 + 2y 2 + 1 = 2(x2 + y 2 ) + 1.
For the first term 2(x2 + y 2 ), it has the largest possible value 2 · 4 = 8 which can be achieved for every
point on the boundary of R. So the absolute maximum value of f (x, y) on R is 8 + 1 = 9.
(b) (0.5 points) Find the absolute minimum value of f (x, y) on R.
1
(b)
Answer.
It’s easy to see that (0, 0) is in R and
f (x, y) = 2x2 + 2y 2 + 1 ≥ 0 + 0 + 1 = 1 = f (0, 0),
for all (x, y) on R.
So f (0, 0) = 1 is the absolute minimum value of f (x, y) on R.
4. Let f (x, y) = x2 + xy 2 − 2x + 1, then
(a) (2 points) Find all critical points of f (x, y).
(a)
Answer.
√
√
(0, 2), (0, − 2) and (1, 0).
Let’s compute the first order partial derivatives of f (x, y), we have
fx (x, y) = 2x + y 2 − 2,
and fy (x, y) = 2xy.
For critical points, we need to solve the following system:
2x + y 2 − 2 = 0,
(1)
2xy = 0
(2)
The equation (2) is much simpler than equation (1), so let’s solve (2) first. Since 2xy = 0, then either
x = 0 or y = 0.
√
√
2
When x = 0, plug√x = 0 into (1),
√ we get y − 2 = 0, then either y = 2 or y = − 2. So we have two
critical points (0, 2) and (0, − 2).
When y = 0, plug y = 0 into (1), we get 2x − 2 = 0, then x = 1. So we have one critical point (1, 0).
In summary, we have three critical points:
√
√
(0, 2), (0, − 2), and (1, 0).
(b) (1 point) Compute the Hessian matrix of f (x, y).
2 2y
2y 2x
(b)
Answer.
.
Since fx (x, y) = 2x + y 2 − 2 and fy (x, y) = 2xy, then
fxx (x, y) = 2,
fxy (x, y) = fyx (x, y) = 2y,
and fyy (x, y) = 2x.
So the Hessian matrix of f (x, y) is:
2 2y
2y 2x
.
(c) (1 point) Compute the discriminant D(x, y) of f (x, y).
4x − 4y 2 .
(c)
Answer.
Since
fxx (x, y) = 2,
fxy (x, y) = fyx (x, y) = 2y,
and fyy (x, y) = 2x,
then
D(x, y) = fxx fyy − (fxy )2 = 2 · (2x) − (2y)2 = 4x − 4y 2 .
(d) (2 points) Classify all critical points of f (x, y). (Make a table)
Answer.
Since D(x, y) = 4x − 4y 2 , then
√
D(0, 2) = −8,
√
D(0, − 2) = −8,
and D(1, 0) = 4.
Since fxx (x, y) = 2, then
(x,√y)
(0, √2)
(0, − 2)
(1, 0)
D(x, y)
−8
−8
4
fxx (x, y)
2
2
2
Conclusion
saddle point
saddle point
local minimum point
Your Score:
/11