MATH 115 MIDTERM 1 SOLUTIONS
Question 1: The normal line to the surface x2 + 3y 2 − z 2 = 3 at the point P0 = (1, 1, 1)
crosses the x, y-plane at the point.
A.
B.
C.
D.
E.
F.
G.
H.
(1,1,1)
(1,1,0)
(2,4,0)
(3,7,0)
(2,6,0)
(0,-2,0)
(0,0,1)
(0,0,2)
Answer: C. (2,4,0)
The gradient of a function that defines a surface is always normal to the surface itself.
Thus, the normal line to the surface x2 + 3y 2 − z 2 = 3 at P0 = (1, 1, 1) would be the point,
P0 , plus all scalar multiples of the gradient at that point. Here, we have:
f (x, y, z) = x2 + 3y 2 − z 2 = 3
so then
∇f = 2xi + 6yj − 2zk
which evaluates to
∇f = 2i + 6j − 2k
at the point P0 = (1, 1, 1). Thus, the normal line to the surface through the point P0 is
given by (1 + 2t)i + (1 + 6t)j + (1 − 2t)k for all t ∈ R (parametrically, this can be written as
x = 1 + 2t, y = 1 + 6t, z = 1 − 2t). This line crosses the x, y plane when the k-component
of the line is 0, or equivalently if z = 1 − 2t = 0. This occurs when t = 21 . At that
point, we have that the x and y values of the point are given by x = 1 + 2( 12 ) = 2 and
y = 1 + 6( 21 ) = 4. Thus, (2,4,0) is the answer.
Date: October 1, 2015.
1
2
MATH 115 MIDTERM 1 SOLUTIONS
Question 2: Find the minimum value of the function f (x, y) = − 23 x − 2y + 1 when
subject to the constraint x2 + y 2 = 1.
A.
B.
C.
D.
E.
F.
G.
H.
2/5
1
17/10
7/2
−9/2
−3/2
−14/10
−13/10
Answer: F. −3/2
Since the substitution method for this question would involve the use of square roots, the
easier approach would be to solve via the Lagrangean. Setting up the Lagrange function,
we have:
3
L(x, y, λ) = − x − 2y + 1 − λ(x2 + y 2 − 1)
2
which gives us the following set of partial derivatives
3
∂L
∂L
∂L
= − − 2λx,
= −2 − 2λy,
= x2 + y 2 − 1
∂x
2
∂y
∂λ
Setting the partials to 0 and then solving for them, we find that
∂L
3
3
= − − 2λx = 0 =⇒ λ = −
∂x
2
4x
3
(since x 6= 0, otherwise ∂L
∂x = − 2 6= 0). Plugging this into the following equation, we find
3y
4
∂L
= −2 − 2λy = −2 +
= 0 =⇒ y = x
∂y
2x
3
Finally, plugging y = 43 x into the final equation, we find that
4
16
25
3
1 = x2 + y 2 = x2 + ( x)2 = x2 + x2 = x2 = 1 =⇒ x = ±
3
9
9
5
3 4
3
4
Thus, our critical points are ( 5 , 5 ) and (− 5 , − 5 ). Checking these values in our original
function, we find that the minimum occurs at ( 35 , 45 ), which gives us the value −3/2.
MATH 115 MIDTERM 1 SOLUTIONS
Question 3: Evaluate
A.
B.
C.
D.
E.
F.
G.
H.
ey
x y
R1R1
0
3
dy dx
0
1
−1
e
−e
e−1
e2
e2 − 1
Answer: F. e − 1
The region over which you’re integrating is when x goes from 0 to 1, y goes from x to 1.
This means it looks like this:
y
(0,1)
(1,1)
(0,0)
x
Equivalently to what you’re given in the problem, you can say that as y goes from 0 to 1,
x goes from 0 to y. In other words, you have that
Z 1Z y y
Z 1Z 1 y
e
e
dy dx =
dx dy
x y
0
0
0 y
You find that, after evaluating the inner integral, you have
Z 1h i
Z 1
h i1
y
y
e x dy =
ey dy = ey = e − 1
0
0
0
0
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MATH 115 MIDTERM 1 SOLUTIONS
Question 4: Find the absolute minimum of the function f (x, y) = x2 − 4x + y 2 − 6y + 3
on the closed triangle bounded by the lines x = 0, y = 4, y = x.
A.
B.
C.
D.
E.
F.
G.
H.
−10
−1
1
3
7/3
18/5
12/5
5
Answer: A. −10
We’re looking to find the absolute minimum of the function f (x, y) = x2 − 4x + y 2 − 6y + 3
over the following area:
y
(0,4)
x=0
(0,0)
y=4
(4,4)
y=x
x
To find the absolute minimum of this function, first check the critical points of the
function. These points occur when the partial derivatives of f are 0, namely when
∂f
= 2x − 4 = 0
∂x
and when
∂f
= 2y − 6 = 0
∂y
The only critical point of this function is x = 2, y = 3. By the second derivative test, we
2
2
can check to see whether this value is a maximum or minimum by evaluating ∂∂xf2 , ∂∂yf2 , and
MATH 115 MIDTERM 1 SOLUTIONS
∂2f
∂x∂y .
5
These evaluate to:
∂2f
= 2,
∂x2
We now find D, which is given below:
D=
∂2f
= 2,
∂y 2
∂2f
=0
∂x∂y
∂2f 2
∂2f ∂2f
−
(
) =2∗2−0=4>0
∂x2 ∂y 2
∂x∂y
2
Since D > 0 and ∂∂xf2 > 0, we know that we have found a local minimum. The value at
this point is f (2, 3) = −10. Now we must check the boundary to see if it is the absolute
minimum.
On the boundary x = 0, then we have f (0, y) = y 2 − 6y + 3. The minimum value of
this function occurs when f 0 (0, y) = 2y − 6 = 0, so when y = 3 (you can check this is a
minimum by noting f 00 (0, y) = 2). At the point (0,3), f (0, 3) = 9 − 18 + 3 = −6.
On the boundary y = 4, we have f (x, 4) = x2 − 4x − 5. This function is minimized when
f 0 (x, 4) = 2x − 4 = 0, so when x = 2 (again, you can check that f 00 (x, 4) = 2, so x = 2 is a
minimum). At the point (2,4), we have f (2, 4) = 4 − 8 − 5 = −9.
On the boundary x = y, we have f (x, x) = 2x2 − 10x + 3. This function is minimized
50
6
−19
when f 0 (x, x) = 4x−10, so when x = 52 . At that point, f ( 52 , 52 ) = 25
2 − 2 + 2 = 2 = −9.5.
We also check the vertices of the triangle, namely the points (0,0), (0,4) and (4,4). At
those points, we have f (0, 0) = 3, f (0, 4) = −5 and f (4, 4) = −5. Thus, the minimum
value on the triangle occurs when (x, y) = (2, 3).
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MATH 115 MIDTERM 1 SOLUTIONS
Question 5: Find the volume of the region that lies under the paraboloid z = 3x2 + 3y 2
and above the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xy plane.
A.
B.
C.
D.
E.
F.
G.
H.
33
2π
27
3π
3/2
0
3/5
4
Answer: H. 4
To find the area of the given paraboloid, you need to integrate the function z = 3x2 + 3y 2
over the given region. That region looks like:
y
(0,2)
(1,1)
x
(0,0)
As you can see from the picture, as x goes from 0 to 1, y goes from x to 2 − x. Thus,
the integral you’re looking at is:
Z 1 Z 2−x
(3x2 + 3y 2 ) dy dx
0
x
Evaluating the integral, we find that we now have
Z 1h
Z 1
i2−x
2
3
3x y + y
dx =
(3x2 (2 − x) + (2 − x)3 ) − (3x3 + x3 ) dx
y=x
0
0
By the binomial theorem, we have that (2 − x)3 = 8 − 12x + 6x2 − x3 , so we can plug this
in above to find the above equals:
Z 1
−8x3 + 12x2 − 12x + 8 dx
0
which evaluates to
h
i1
− 2x4 + 4x3 − 6x + 8x = −2 + 4 − 6 + 8 = 4
0
MATH 115 MIDTERM 1 SOLUTIONS
7
Question 6: A cylindrical tin can (with a top and a bottom) is to be built out of 6π
square inches of tin. What is the maximum volume of such a can?
A. 9π
4 cubic inches
B. 3π cubic inches
C. 27π
8 cubic inches
D. 9π
2 cubic inches
E. π cubic inches
F. 5π
4 cubic inches
G. 2π cubic inches
H. 5π
2 cubic inches
Answer: G. 2π cubic inches
We know that the surface area of a cylinder is given by SA = 2πrh + 2πr2 and that the
volume of the cylinder is given by V = πr2 h. Thus, we are looking to maximize V subject
to SA = 6π, which can be equivalently written as rh + r2 = 3 by dividing by 2π on both
sides. From here, you can either maximize using the Lagrange method or via substitution.
To use the substitution method, note that r 6= 0 because then the volume would be 0.
2
Thus, we can rewrite the constraint rh + r2 = 3 as h = 3−r
r . Substituting this into the
volume equation, we find that
3 − r2
= 3πr − πr3
r
. Taking the derivative of V with respect to r, we find that 3π − 3πr2 = 0 =⇒ r2 = 1.
2
Thus, r = 1 since r > 0. We know this is a maximum value because ddrV2 = −6πr < 0 if
2
r > 0. At this value, we have that h = 3−1
1 = 2. So V = π(1)(2) = 2π.
V = πr2
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MATH 115 MIDTERM 1 SOLUTIONS
Question 7: Let f (x, y) = 2x2 ey +1. Use the differential to approximate 2(3.05)2 e0.1 +1.
A.
B.
C.
D.
E.
F.
G.
H.
21.4
3.4
21.1
3.1
9.4
9.1
21.2
22.29
Answer: A. 21.4
The differential is used to estimate a small change in f around a given point that you
know. Here, we know that 2 ∗ (3)2 e0 + 1 = 2 ∗ 9 + 1 = 19. Thus, we can estimate
2(3.05)2 e0.1 + 1 by using the differential to measure how much a small change in x and y
affects the value of f .
The differential formula is given by df = fx dx + fy dy. Here, fx = 4xey and fy = 2x2 ey .
Thus, in the neighborhood of the point (3,0), we have that df = 4xey dx+2x2 ey dy = 12dx+
18dy. Since we are looking to approximate f (3.05, 0.1), we have that dx = 3.05 − 3 = 0.05
and dy = 0.1 − 0 = 0.1. So, df = 12(0.05) + 18(0.1) = 0.6 + 1.8 = 2.4. Since f (3, 0) = 19,
then f (3.05, 0.1) is about 19 + 2.4 = 21.4.
MATH 115 MIDTERM 1 SOLUTIONS
9
Question 8: Consider the function f (x, y) = x2 + 2xy − y 2 − 10x + 2y. Find its critical
points and determine their type.
A.
B.
C.
D.
E.
F.
G.
H.
Local min at (2,3)
Saddle point at (2,3)
There are no critical points
There is more than one critical point
Local max at (0,0)
Local min at (0,0)
Saddle point at (0,0)
Local max at (2,3)
Answer: B. Saddle point at (2,3).
The critical points of the function occur when fx = 0 and fy = 0. This occurs when
fx = 2x + 2y − 10 = 0 and when fy = 2x − 2y + 2 = 0. We can rewrite the first equation
as 2y = 10 − 2x and the second as 2y = 2x + 2 so we have that 10 − 2x = 2x + 2, which
means x = 2 so y = 3.
To determine whether this point is a min, a max, or a saddle point, we must apply the
second derivative test. To do this, we look at D = fxx fyy − (fxy )2 . Here, fxx = 2, fyy =
−2, fxy = 2. So we have that D = 2(−2) − (2)2 = −4 − 4 = −8 < 0. Thus, we have that
the point (2,3) is a saddle point of our given function.
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MATH 115 MIDTERM 1 SOLUTIONS
Question 9: Consider the function f (x, y) = 2uv + 2ev−1 − 2 cos(u) + 1 where u and v
are functions of x and y with the properties:
(1) u(2, −1) = 0
(2) v(2, −1) = 1
(3) ∂u
∂x (2, −1) = 2
∂u
(4) ∂y (2, −1) = 3
∂v
(5) ∂x
(2, −1) = 5
∂v
(6) ∂y (2, −1) = 7
Then
A.
B.
C.
D.
E.
F.
G.
H.
∂f
∂x
at the point (x, y) = (2, −1) is:
−4 + 2e−2 − 2 cos(2)
14
18
32
0
2
-2
2e−1 − 2
Answer: B. 14
By the chain rule, you have that
∂f ∂u ∂f ∂v
∂f
=
+
∂x
∂u ∂x ∂v ∂x
where
∂f
∂f
= 2v + 2 sin(u),
= 2u + 2ev−1
∂u
∂v
At (x,y) = (2,1), we have that
∂f
= 2 ∗ v(2, −1) + 2 sin(u(2, −1)) = 2 ∗ 1 + 2 ∗ sin(0) = 2
∂u
and
∂f
= 2 ∗ u(2, −1) + 2 ∗ ev(2,−1)−1 = 2 ∗ 0 + 2 ∗ e1−1 = 2
∂v
Thus, we have that
∂f
∂u
∂v
=2∗
(2, −1) + 2 ∗
(2, −1) = 2 ∗ 2 + 2 ∗ 5 = 14
∂x
∂x
∂x
Thus, the answer is 14.