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Unformatted text preview: 2 Practice Exam 2 The multichoice answers are all A.
Problem (#11). Let f (x, y ) = x2 y + xy 2 + 3xy . Find the critical points of
f and tell what type each one is.
Solution. A point (x0 , y0 ) is a critical point of f just in case either ∇f (x0 , y0 )
is undeﬁned or ∇f (x0 , y0 ) = 0. We compute,
∇f = 2xy + y 2 + 3y, x2 + 2xy + 3x This is deﬁned for all (x, y ), so we only need to worry about the points where
the gradient is 0. For these, we solve,
2xy + y 2 + 3y = 0
2xy + x2 + 3x = 0
Equivalently, we can solve the system:
y (2x + y + 3)
x(x + 2y + 3) =0
=0 From the ﬁrst equation, we see that y = 0 or 2x + y + 3 = 0. We then derive
y = 0 =⇒ x(x + 2 · 0 + 3) = 0 =⇒ (x = 0 or x = −3)
(2x + y + 3 = 0 and x = 0) =⇒ y = −3
2x + y + 3 = 0
(2x + y + 3 = 0 and x = 0) =⇒
=⇒ x = y = −1
x + 2y + 3 = 0
Thus, the critical points of f are (0, 0), (−3, 0), (0, −3), (−1, −1).
To classify them, we use the second derivative test. First, note that
fxx fxy
Df (x, y ) = det
fyx fyy
2y
2x + 2y + 3
= det
2x + 2 y + 3
2x
= 4xy − (2x + 2y + 3)2 • Df (0, 0) = −9 < 0, so (0, 0) is a saddle point
4 • Df (−3, 0) = −9 < 0, so (−3, 0) is a saddle point
• Df (0, −3) = −9 < 0, so (0, −3) is a saddle point
• Df (−1, −1) = 3 > 0 and fxx (−1, −1) = −2 < 0, so (−1, −1) is a local
maximum point. Problem. Suppose that (1, 1) is the only critical point of the function f (x, y ) =
2x − 2xy + y 2 + 1. Find the absolute maximum value of the function on the
rectangle R = [0, 2] × [0, 3].
Solution. To ﬁnd the absolute maximum of f over R, we ﬁnd the critical
points of f (which is done for us, here) and ﬁnd the local maximum points
of f on the boundary; then we compare the values of f at these points.
The boundary of R has (unfortunately) 4 parts.
• L1 = {(x, y ) : y = 0, 0 ≤ x ≤ 2}
• L2 = {(x, y ) : y = 3, 0 ≤ x ≤ 2}
• L3 = {(x, y ) : x = 0, 0 ≤ y ≤ 3}
• L4 = {(x, y ) : x = 2, 0 ≤ y ≤ 3}
We will ﬁnd the local maximum points of f on each of L1 through L4 , and
we will do this by parametrizing each segment.
. L1 : Maximize f (x, y ) subject to y = 0, 0 ≤ x ≤ 2.
Take x(t) = t, y (t) = 0. Then the stated problem is the same as ﬁnding
the maximum of g1 (t) = f (x(t), y (t)) for t in [0, 2]. Now, g1 (t) = 2t + 1,
and the maximum occurs when t = 2. Hence, maximum point of f (x, y )
restricted to L1 is (2, 0).
. L2 : Maximize f (x, y ) subject to y = 3, 0 ≤ x ≤ 2.
Take x(t) = t, y (t) = 3. Again, the stated problem is the same as
ﬁnding the maximum of g3 (t) = f (x(t), y (t)) for t in [0, 2]. Now, g2 (t) =
2t − 6t + 32 + 1 = −4t + 10, and the maximum occurs when t = 0. Hence,
maximum point of f (x, y ) restricted to L2 is (0, 3).
. L3 : Maximize f (x, y ) subject to x = 0, 0 ≤ y ≤ 3. 5 Take x(t) = 0, y (t) = t. Then the stated problem is the same as ﬁnding
the maximum of g3 (t) = f (x(t), y (t)) for t in [0, 3]. Now, g3 (t) = t2 + 1,
and the maximum occurs when t = 3. Hence, maximum point of f (x, y )
restricted to L3 is (0, 3).
. L4 : Maximize f (x, y ) subject to x = 2, 0 ≤ y ≤ 3.
Take x(t) = 0, y (t) = t. Then the stated problem is the same as ﬁnding
the maximum of g3 (t) = f (x(t), y (t)) for t in [0, 3]. Now, g3 (t) = 3 − 6t +
t2 + 1 = t2 − 6t + 1. Note that g4 (t) = 2t − 6, and the maximum occurs
when t = 0. Hence, maximum point of f (x, y ) restricted to L3 is (2, 0).
Finally, we compare the values of f at the points we’ve found:
• f (1, 1) = 2
• f (2, 0) = 5
• f (0, 3) = 10
So, the maximum value of f over R is 10, and this occurs at (0, 3).
Problem. Find the maximum and minimum values of f (x, y, z ) = 2x − z
subject to x2 + 10y 2 + z 2 = 5.
Solution. Setting g (x, y, z ) = x2 + 10y 2 + z 2 , we see that we need to solve
the system,
∇f = λ∇g
x2 + 10y 2 + z 2 = 5
This expands to 2 0 −1 2
x + 10y 2 + z 2 = 2λ x
= 20λy
= 2λz
=5 To solve this, ﬁrst note that λ = 0, for if λ = 0, we’d get 2 = 0 in the ﬁrst
equation. Since λ = 0 and 20λy = 0, we now know that y = 0.m Also, λ = 0
means that we can divide by λ, so from the ﬁrst and third equations we get
1
1
x = λ and z = −λ . Substituting all of this into the last equation, we get
2
2
2
1
2 −1
+ 10 · 0
=5
λ
2λ
5
so 4λ2 = 5, and so λ = ± 1 . If λ = 1 , we get x = 2, y = 0, z = −1, and if
2
2
λ = − 1 , we get x = −2, y = 0, z = 1. Thus,
2 6 • f (2, 0, −1) = 5 is the maximum of f on the surface x2 + 10y 2 + z 2 = 5
• f (−2, 0, 1) = −5 is the minimum of f on the surface x2 +10y 2 + z 2 = 5 7 ...
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This note was uploaded on 11/16/2011 for the course MULTIVARIA 251 taught by Professor Staff during the Fall '11 term at Rutgers.
 Fall '11
 Staff

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