Solutions to exam 2 practice problems

Math 241H 0101 Exam 2 Practice Problems
Exam 2 will cover Chapter 13.
1. For z = f(x, y) = xy, sketch and label the level curves for c = ±1, ±2.
For the level curve c = 1 you should have the graph of the function y = 1/x. For the
level curve c = −1 you should have the graph of the function y = −1/x. For the level
curve c = 2 you should have the graph of the function y = 2/x. For the level curve
c = −2 you should have the graph of the function y = −2/x.
2. Let
g(x, y) =
2x − y2
2x + y 2 .
Does the following limit exist? If so, what is the limit? If not, why not?
lim g(x, y).
(x,y)→(0,0)
The limit does not exist. If we consider points (x, 0) along the positive x-axis approaching
(0, 0), g(x, 0) = 1. On the other hand, if we consider points (0, y) along the
positive y-axis approaching (0, 0), g(0, y) = −1.
3. Let f(x, y) = 3x 2 + 6xy − y 3 .
Find the directional derivative of f(x, y) in the direction of 3⃗i + ⃗j at (1, 1).
First we want to find a unit vector pointing in the same direction as 3⃗i + ⃗j.
⃗u = 3 √
10
⃗i + 1 √
10
j.
Then we want to compute partial derivatives:
f x (1, 1) = 6x + 6y
∣ = 12
(1,1)
∣
∣∣∣(1,1)
f x (1, 1) = 6x − 3y 2 = 3
1

Then
D ⃗u f(1, 1) = 36 √
10
+ 3 √
10
= 39 √
10
.
4. A differentiable function f(x, y) has the property that f(1, 3) = 7 and ∇f(1, 3) =
2⃗i − 5⃗j.
(a) Find the equation of the tangent line to the level curve of f through the point
(1, 3).
To find the equation of the tangent line we need the slope of the tangent line and
a point on the line. We know that (1, 3) is a point on the line. We know the
vector 2⃗i − 5⃗j is perpendicular to the line. Then the vector 5⃗i + 2⃗j is tangent to
the line since the dot product of the two vectors is zero.
So the equation for the tangent line is:
y = 2 (x − 1) + 3.
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(b) Find the equation of the tangent plane to the surface z = f(x, y) at the point
(1, 3, 7).
Let g(x, y, z) = f(x, y) − z. Then the surface z = f(x, y) is the level surface of
g(x, y, z) when c = 0. Then, ∇g = f x
⃗i + f y
⃗j − ⃗ k is a vector perpendicular to the
tangent plane. So the equation of the plane is:
2(x − 1) − 5(y − 3) − (z − 7) = 0.
5. Let F (u, v) be a function of two variables. Find f ′ (x) if f(x) = F (x, x).
Using the chain rule we get
f ′ (x) = ∂F ∂u
∂u ∂x + ∂F ∂v
∂v ∂x
= F u (x, x) + F v (x, x)
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6. Let f(x, y) = sin(xy).
(a) What is the direction in which f increases most rapidly at the point (1, π).
∇f(1, π) = f x (1, π)⃗i + f y (1, π)⃗j
= y cos(xy)
∣
⃗i + x cos(xy)
∣
⃗j
(1,π) (1,π)
= −π⃗i − ⃗j
(b) What is the maximal directional derivative at the point (1, π).
||∇f(1, π)|| = √ π 2 + 1
7. Let f(x, y) = sin(x + 2y). Estimate f(π/2 + 0.01, π/4 − 0.01).
First compute partial derivatives:
f x (π/2, π/4) = cos(x + 2y)
∣ = cos(π) = −1
(π/2,π/4) f y (π/2, π/4) = 2 cos(x + 2y)
∣ = 2 cos(π) = −2
(π/2,π/4)
The change in f is approximately
∆f ≈ f x (π/2, π/4)∆x + f y (π/2, π/4)∆y
= −1(0.01) − 2(−0.01) = 0.01
Since f(π/2, π/4) = 0, f(π/2 + 0.01, π/4 − 0.01) ≈ 0.01.
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8. Let f(x, y) = x 3 + y 3 − 6y 2 − 3x + 9. Find and classify the critical points as local
minima, local maxima, saddle points or neither.
First compute the gradient vector:
∇f = (3x 2 − 3)⃗i + (3y 2 − 12y)⃗j.
The gradient always exists so the critical points are only when the gradient is zero.
This happens when x = ±1 and y = 0, 4. So we have four critical points to examine:
(−1, 0), (−1, 4), (1, 0) and (1, 4).
Compute the second derivatives:
f xx = 6x
f xy = 0
f yy = 6y − 12
Then, the discriminant is:
D = f xx f yy − f 2 xy.
For each critical point compute the discriminant:
D(−1, 0) = (−6)(−12) − 0 = 72 > 0
D(−1, 4) = (−6)(12) − 0 = −72 < 0
D(1, 0) = 6(−12) − 0 = −72 < 0
D(1, 4) = 6(12) − 0 = 72 > 0
So by the second derivative test, (−1, 0) is a local maximum, (−1, 4) and (1, 0) are
saddle points and (1, 4) is a local minimum.
9. Find the maximum and minimum values of f(x, y) = x 2 + 3y 2 on the quarter circle
x 2 + y 2 ≤ 4 with x, y ≥ 0.
First solve for the critical points:
So we have a critical point at (0, 0).
f x = 2x
f y = 6y
Since f(0, 0) = 0 and for all other points (x, y) on the region f(x, y) ≥ 0, 0 is a global
minimum value of f.
We know that there exists a global maximum and minimum since the region is closed
and bounded. Since there are no other critical points, the global maximum must occur
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on the boundary. The boundary has three pieces. The first piece is when x = 0 and
0 ≤ y ≤ 2; along this piece, f achieve its maximum value when y = 2, f(0, 2) = 12.
The second piece is when y = 0 and 0 ≤ x ≤ 2; along this piece, f achieves its
maximum value when x = 2, f(2, 0) = 4.
The last piece is the circular arc, where x 2 + y 2 = 4. Then along this piece, f(x, y) =
4 + 2y 2 . This will achieve its maximum when y is as big as possible, i.e. when y = 2
but then x = 0. So f(0, 2) = 12 is the value of the global maximum.
10. A closed rectangular box has volume 27cm 3 . What are the lengths of the edges giving
the minimum surface area? Don’t forget to justify why it is a global minimum.
Let l, w, h be the length, width and height of the box. Since the volume of the box is
27, l = 27/(wh).
The total surface area is 2wh + 2lh + 2lw. Substituting the relation above we get a
function of two variables:
First solve for the critical points:
f(w, h) = 2wh + 54
w + 54
h .
f w = 2h − 54
w 2 = 0
h = 27
w 2
f h = 2w − 54 = 0
h 2
54
2w −
= 0
(27/w 2 ) 2
w − w4
27
= 0
w(1 − w3
27 ) = 0
w = 0, w = 3
Since w must be nonzero, w = 3 and h = 3 is the only critical point.
To check that it is a local minimum, use the second derivative test:
f ww = 108
w 3
f hw = 2
f hh = 108
h 3
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Then evaluate the discriminant at the point (3, 3),
D = ( 108
27 )2 − 4 = 16 − 4 = 12 > 0
Since D > 0 and f ww > 0, the critical point is a local minimum.
The possible values of w, h are w, h > 0. Notice that f(3, 3) = 54.
To check that it is a global minimum, check that in each of the following situations
f(w, h) > 54.
(a) w ≤ 1/4
(b) h ≤ 1/4
(c) w ≥ 1/4, h ≥ 400
(d) h ≥ 1/4, w ≥ 400
Since the region R where 1/4 ≤ w ≤ 400 and 1/4 ≤ h ≤ 400 is closed and bounded f
achieves a global minimum on R. Since there is only one critical point and the value
at that critical point is less than the values of f along the entire boundary the critical
point is a global minimum.
11. Let f(x, y) = x 2 + 6y 2 and let g(x, y) = x + 3y.
(a) Use Lagrange multipliers to find the minimum of f(x, y) subject to the constraint
g(x, y) = 10. Assume that there is a global minimum.
(b) Explain why f(x, y) has no maximum subject to the constraint g(x, y) = 10.
(c) Explain why g(x, y) has a maximum and a minimum subject to the constraint
f(x, y) = 9.
(d) Find the maximum and minimum of g(x, y) subject to the constraint f(x, y) = 9.
(a) Notice I added ”Assume that there is a global minimum” to the question. If we assume
this, then we know it must occur at a point where either the gradients of f and g are
parallel, an endpoint of the constraint or when the gradient of g is zero.
First compute the gradients of f and g :
∇f(x, y) = 2x⃗i + 12y⃗j
∇g(x, y)
= ⃗i + 3⃗j
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Since the gradient of g is never zero, and g = 10 has no endpoints, set the gradients
parallel to each other
∇f(x, y) = λ∇g(x, y)
This becomes
2x = λ 12y = 3λ
2x = λ 4y = λ
So, 2x = 4y or x = 2y. Plug this into our constraint equation x + 3y = 10 and get
y = 2. So x = 4.
So the minimum is (4, 2).
(b) f(x, y) has no maximum because we can make x as large as we want and still satisfy
the constraint. This will make f(x, y) arbitrarily large.
(c) g(x, y) has a max and min subject to the constraint f(x, y) = 9 because that constraint
is described by a closed curve. By the extreme value theorem, g achieves a max and
min over that curve.
(d) First compute the gradients of f and g :
∇f(x, y) = 2x⃗i + 12y⃗j
∇g(x, y)
= ⃗i + 3⃗j
Since the gradient of f is zero only at the origin (which does not lie in our constraint),
and the constraint has no endpoints, set the gradients parallel to each other
λ∇f(x, y) = ∇g(x, y)
This becomes
λ2x = 1 12yλ = 3
2x = 1/λ 4y = 1/λ
So, 2x = 4y or x = 2y. Plug this into our constraint equation (2y) 2 + 6y 2 = 9.
This gives y = ±3 √
10
. This yields the following two points: ( 6 √
10
,
3 √
10
) and ( −6 √
10
, −3 √
10
).
Since g( 6 √
10
,
3 √
10
) = 15 √
10
and g( −6 √
10
, −3 √
10
) = −15 √
10
the first point is where the global max
occurs and the second point is where the global min occurs.
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