Quiz 3
Time: 16 minutes
Friday, 22 Jan.
Name
Student number :
Note: Each wrong answer has 13 negative score and each right answer has 1 positive
score. If you are not sure about a question, leave it blank.
1. Assume the vector hm, n, 1i is parallel to both planes 2x−y = 5 and x+z = y −1.
What is m+n?
a) −1
b) 0
c) 3
d) 16
2. Assume Q is the plane passing through the point P (1, 2, 6) that is orthogonal to
the vector h4, 0, −1i. Which point lies on Q?
a) ( 21 , 0, 0)
b) (0, 2, 0)
c) (0, 0, 2)
d) (1, 1, 1)
3. Consider the surface z 3 x2 = z + 4z 2 + y 2 . The point (2, 1) lies on which level
curve of this surface?
a) z = −1
b) z = 0
c) z = 1
d) z = 2
4. Which plane is not parallel to the vector h−1, 1, −1i
a) x = z
b) x = y
x+y =1
d) 2x + y − z = 20
∂2f
(−1, 1).
∂y∂x
+1
d) 2e − 1
5. Let f (x, y) = exy y 2 + xy . Evaluate
a)
4
e
+1
b)
−4
e
−1
c)
2
e
6. How many saddle points does the function f (x, y) = x3 + y 3 + 3x2 − 3y 2 − 8 have?
a) none
b) one
c) two
d) three
7. According to market research, the demand curve for a local pizza restaurant
satisfies the following relation: if p is the price of a pizza (in dollars), and q is the
number of pizzas sold per day, then p2 + 4q 2 = 800. If you are the owner of this
restaurant, what price should you charge for each pizza in order to make your
daily revenue as high as possible?(use
Lagrange Multipliers)
√
a) 200
b) 10
c) 20 2
d) 20
8. Use the method of Lagrange multipliers to find the minimum value of f (x, y) =
(x + 1)2 + (y − 2)2 on the circle x2 + y 2 = 125.
a) 40
b) 80
c) 160
d) 180
9. Find the absolute maximum of g(x, y) = 2x2 − 4x + y 2 − 4y + 1 on the closed
triangular plate with vertices (0, 0), (2, 0) and (2, 2).
a) -1
b) 0
c) 1
d) 2
Quiz 3
Time: 16 minutes
Friday, 22 Jan.
10. Whatis the absolute maximum of f (x, y) = x2 + xy + 3x + 2y + 2 on the domain
D = (x, y) ∈ R2 : x2 ≤ y ≤ 4 ?
a) 16
b) 28
c) 36
d)40
Quiz 3
Time: 16 minutes
Friday, 22 Jan.
Solutions:
Problem 1 : (c) The vector hm, n, 1i has to be orthogonal to normal vectors of
both planes. So
hm, n, 1i . h2, −1, 0i = 2m − n = 0
hm, n, 1i . h1, −1, 1i = m − n + 1 = 0
Hence n = 2 and m = 1.
Problem 2 : (c) The equation of the plane is 4x−z = −2. By pluging in the points,
we see that only the point (0, 0, 2) lies on Q.
Problem 3 : (a) By setting x = 2 and y = 1 in the surface’s equation, we get
4z 2 = z + 4z 2 + 1, so z = −1.
Problem 4 : (b) We need to find the plane whose normal vector is orthogonal to
the vector h−1, 1, −1i.
Problem 5 : (d)
∂f
∂x
= yexy y 2 +
1
y
= exy y 3 +
1
y
and then
∂2f
1
= xexy y 3 + 3exy y 2 − 2 .
∂x∂y
y
Therefore
∂2f
∂x∂y (−1, 1)
= 2e−1 −
1
1
=
2
e
− 1.
Problem 6 : (c)
fx = 3x2 + 6x ⇒ x = 0, −2
fy = 3y 2 − 6y = 0 ⇒ y = 0, 2
So there are four critical points (−2, 0), (−2, 2), (0, 0) and (0, 2). Now we compute
the second partial derivatives of f:

 fxx = 6x + 6
fxy = 0

fyy = 6y − 6
2 = 36(x + 1)(y − 1). By plugging in, we see that
Therefore D(x, y) = fxx fyy − fxy
D(−2, 0) = 36, D(−2, 2) = −36, D(0, 0) = −36 and D(0, 2) = 36. Only (−2, 2)
and (0, 0) are saddle.
Quiz 3
Time: 16 minutes
Friday, 22 Jan.
Problem 7 : (d) We have to maximize the revenue function R(p, q) = pq subject
to g(p, q) = p2 + 4q 2 − 800 = 0. We write
∇R(p, q) = λ∇g(p, q) ⇒ hq, pi = λ h2p, 8qi = h2λp, 2λqi .
So we need to solve the system of equations

(1)
 q = 2λp
p = 8λq
(2)
 2
p + 4q 2 = 800 (3)
If we plug in q = 2λp in the last equation, we get (1 + 16λ2 )p2 = 800, so p 6= 0
q
and therefore λ 6= 0. Now write, using (1), λ = 2p
and plug in the value of λ in
2
q
(2). Then one get p = 8 2p
q = 4qp , so q = 12 p (q = − 21 p is not acceptable as p
stands for price). Now we plug in this in the last equation to get p2 + p2 = 800,
so p = 20.
Problem 8 : (b) We need to minimize f (x, y) = (x + 1)2 + (y − 2)2 subject to
g(x, y) = x2 + y 2 = 125. Following Lagrange Multipliers method, we need to
solve
∇f (x, y) = λ∇g(x, y)
g(x, y) = x2 + y 2 = 125
Since ∇f (x, y) = h2(x + 1), 2(y − 2)i and ∇g(x, y) = h2x, 2yi, the system simplifies to:

 2(x + 1) = λ∇(2x) (1)
2(y − 2) = λ∇(2y) (2)

g(x, y) = x2 + y 2 = 125 (3)
If first try to eliminate λ from the first two equations and get another relation
between x and y. We would like to write λ = 2(x+1)
= λ 2(y−2)
2x
2y . But the denominators may be zero for some x or/and y. So we separate different cases:
Case 1: If x = 0, from (1) we get: 2 = 0 which is impossible and we don’t get any
point here.
Case 2: If y = 0, from (2) we get −2 = 0 which is again impossible and we don’t
get any point in this case neither.
Case 3! If x 6= 0 and y 6= 0, then we can finally write, using (1) and (2):
λ=
2(x + 1)
2(y − 2)
=λ
(4),
2x
2y
Quiz 3
Time: 16 minutes
Friday, 22 Jan.
After simplifying (4), we get
y = −2x (5),
and putting this in (3), we get x2 + (−2x)2 = 125 which gives two values for x:
x1 = 5 and x2 = −5. For x1 = 5 we get, from (5), y1 = −10 (and λ = 56 ) ,and for
x2 = −5 we get y2 = 10 (and λ = − 45 ).
Now we check the values of f at these two points:
f (5, −10) = 62 + (−12)2 = 180f (−5, 10) = (−4)2 + 82 = 80.
80 is the minimum of f . (and 180 is its maximum)
Note: After finding x1 (or x2 ), if you try to find y1 , using the equation (3), you
get two solutions y1 = −10 and y10 = 10. But the second one is not valid, as if you
plug in x1 = 5 and y10 = 10 in (1), you get λ = 56 and from (2), you get λ = 45 . So
we don’t get a triple (x, y, λ) for solution of the above system of equations. This
is why it is better to use (5) rather than (4).
Problem 9 : (c) First we find the critical points of f :
fx = 4x − 4 = 0
fy = 2y − 4 = 0
So there is only one critical point (1, −2). But this point is not inside or on the
boundary of the triangle and therefore we ignore it. Now we have to parametrise
each side of the triangle:
The side connecting (0,0) to (2,0): C1 = {(x, 0) : 0 ≤ x ≤ 2}. We define and
simplify the function g1 as:
g1 = f (x, 0) = 2x2 − 4x + 02 − 4 ∗ 0 + 1 = 2x2 − 4x + 1.
(it gives the values of f on C1 ) Now we find the critical points of g1 in the interval
x ∈ [0, 2]:
g10 (x) = 4x − 4 = 0 ⇒ x = 1 Then g1 (1) = f (1, 0) = −1. We also have to take
into account the values of g1 at endpoints of the interval [0, 2]:
g1 (0) = f (0, 0) = 1 and g1 (2) = f (0, 2) = 1.
The side connecting (0,0) to (2,2): C2 = {(x, x) : 0 ≤ x ≤ 2} and
g2 (x) = f (x, x) = 3x2 − 8x + 1.
g20 (x) = 6x − 8 = 0 ⇒ x = 34 . Then g2 ( 43 ) = −13
3 . We also need to consider
g2 (0) = f (0, 0) = 1 and g2 (2) = f (2, 2) = −3.
The side connecting (2,0) to (2,2): C3 = {(2, y) : 0 ≤ y ≤ 2}. Then define
g3 (y) = f (2, y) = y 2 − 4y + 1
Quiz 3
Time: 16 minutes
Friday, 22 Jan.
So g30 (y) = 2y − 4 = 0 ⇒ y = 2. This gives one of the endpoints of C3 . So only
need to compute g3 (0) = f (2, 0) = 1 and g3 (2) = f (2, 2) = −3. Therefore the
maximum of f is 1 (happening at three points (0, 0), (0, 2) and (2, 0). The mini4 4
mum of f is −13
3 happening at ( 4 , 3 ).
Problem 10 : (b) First, we compute the critical points of f :
fx = 2x + y + 3 = 0
fy = x + 2 = 0
We get only one critical points (−2, 1) which is not in D so we ignore it! Now we
analyse the boundary:
The side connecting y = 4: C1 = {(x, 4) : −2 ≤ x ≤ 2}. Define
g1 (x) = f (x, 4) = x2 + 7x + 10.
Then g10 (x) = 2x + 7 = 0 so x = −7
2 which is not in the interval [−2, 2] and we
have to ignore it. The values of g at endpoints are
g1 (−2) = f (−2, 4) = 0, g1 (2) = f (2, 4) = 28.
The side connecting y = x2 : C1 = {(x, x) : −2 ≤ x ≤ 2}. Define
g2 (x) = f (x, x2 ) = x3 + 3x2 + 3x + 2.
So g20 (x) = 3x2 + 6x + 3 = 3(x2 + 2x + 1) = 3(x + 1)2 = 0 so x = −1 and
g2 (−1) = f (−1, 1) = 1. The values of g at endpoints are
g2 (−2) = f (−2, 4) = 0, g2 (2) = f (2, 4) = 28.
So the maximum of f is 28 and its minimum is 10.