C Roettger, Spring 15
Math 165 - Exam 2C - solutions
Problem 1 A farmer wants to fence in a rectangle. The area is supposed to
be 20,000 square feet. She wants to use sturdy fence, costing $ 9 per linear
foot for the two shorter sides, and cheap fence, costing $ 6 per linear foot,
for one of the longer sides. The other longer side of the rectangle uses a wall
that is already built, so that side costs nothing. Which dimensions of the
rectangle give the lowest cost? Give exact answers and then round to the
nearest integer.
Solution. Let x be the length of the shorter side and y the length of the
longer side of the rectangle, A its area, and C the total cost. Then it is given
that A = 20000. We also have the equations
A = xy
C = 18x + 6y
We use the first equation above with A = 20000 to eliminate y from the
second equation, so we can write C as a function of only one variable, namely
x:
120000
C(x) = 18x +
x
There is no lower limit for x, but since x has to be the shorter side, we get
x ≤ y and therefore
x2 ≤ xy = 20000
√
√
so x2 ≤ 20000 and x ≤ 100 2. The domain for x is then (0, 100 2]. We
proceed
√ to find critical points for in the domain. Besides the endpoint x =
100 2, get them from solving C 0 (x) = 0 with
C 0 (x) = 18 −
120000
x2
so
x2 = 120000/18 =
20000
3
p
√
so x = 100 2/3 ≈ 81 and y = 100 6 ≈ 245 (in feet). This x-value is indeed
in the domain. The sign pattern of C 0 (x) is −+ which means that this last
point really gives the minimum cost.
Problem 2 Find the following antiderivatives. Make sure to list the function
g amd its derivative if you are using Substitution.
Z
1
a)
x4 + 30x2 − 2x + 5 dx = x5 + 10x3 − x2 + 5x + C
5
Z
b)
2 cos(t) + sin(t) dt = 2 sin(t) − cos(t) + C
Z
sin5 (3y)
+C
c)
cos(3y) sin4 (3y) dy =
15
Solution. I used the substitution u(y) = sin(3y) in c), so du = 3 cos(3y) dy.
Problem 3 Consider the function
f (t) = t165 .
a) What is the linearization L(t) of f (t) at t = 1?
b) Use part a) to approximate f (0.98). Exact evaluations of the function f
will not give credit.
Solution. a) We differentiate f ,
f 0 (t) = 165t164
so f 0 (1) = 165. Also, f (1) = 1. So the linearization is
L(t) = 165(t − 1) + 1.
b) This just means using L(t) instead of f (t), which makes sort of sense
because 1.02 is close to 1 (although not very close, given the huge exponent/steep slope of the tangent line). We get
f (0.98) ≈ L(0.98) = 165 · (−0.02) + 1 = dy + 1 = −3.3 + 1 = −2.3.
Problem 4 For the function
f (x) = (x2 − 3)e−2x
a) find all critical points,
b) find the intervals on which f (x) is increasing and the intervals on which
f (x) is decreasing,
c) find all x-values where f (x) has a local minimum,
d) and find all x-values where f (x) has a local maximum.
Solution. In the graph below, note how easy it is to miss the local maximum
with the naked eye.
a) Since f 0 (x) is defined everywhere,
f 0 (x) = (2x − 2x2 + 6))e−2x = −2(x2 − x − 3)e−2x
we just need to solve f 0 (x) = 0. This is true exactly when
x2 − x − 3 = 0
√
so for x = (1 ± 13)/2.
0
b) The
√ sign pattern√of f (x) is − + −, so the function f (x) is increasing in
[(1 − 13)/2, (1 + 13)/2] and decreasing outside of this interval.
√
c), d) From a) and b), f (x) √
has a local minimum at x = (1 − 13)/2 and a
local maximum at x = (1 + 13)/2.
Problem 5 A rocket is launched vertically and tracked by a radar station 5
miles from the launch site. At a moment when the distance of the rocket from
the radar station is 7 miles, this distance increases by 1600 miles per hour.
What is the vertical speed of the rocket at this moment in time? Round the
answer to a multiple of 10.
Solution. First, we introduce names for the relevant quantities. Let z be
the distance of the rocket from the radar station. This is the hypotenuse of a
right triangle, where one leg (horizontal) has length 5 and one leg (vertical)
has length h, the height of the rocket. Then write equations between these
quantities:
z 2 = 52 + h2
and we know z 0 = 1600 at the moment when z = 7. Then we differentiate
the equation above:
2zz 0 = 2hh0
Substitute the numbers given for the particular moment in time. We have
to use z = 7 in the original equation to get
√
√
h = z 2 − 52 = 2 6
√
2 · 7 · 1600 = 4 · 6h0 .
Solving for h0 gives
5600
h0 = √ ≈ 2290
6
(in miles per hour).