C Roettger, Spring 15
Math 165 - Exam 2A - solutions
Problem 1 A farmer wants to fence in a right triangle. The area is supposed
to be 5000 square feet. She wants to use sturdy fence, costing $ 9 per linear
foot for the shorter leg, and cheap fence, costing $ 6 per linear foot, for the
longer leg of the triangle. The hypotenuse of the triangle uses a wall that is
already built, so that side costs nothing. Which dimensions of the triangle
give the lowest cost? Give exact answers and then round to the nearest
integer.
Solution. Let x be the length of the shorter leg and y the length of the
longer leg of the triangle, A its area, and C the total cost. Then it is given
that A = 5000. We also have the equations
xy
2
C = 9x + 6y
A =
We use the first equation above with A = 5000 to eliminate y from the second
equation, so we can write C as a function of only one variable, namely x:
C(x) = 9x +
60000
x
There is no lower limit for x, but since x has to be the shorter leg of the
triangle, we get x ≤ y and therefore
x2
xy
≤
= 5000
2
2
so x2 ≤ 10000 and x ≤ 100. The domain for x is then (0, 100]. We proceed
to find critical points for in the domain. Besides the endpoint x = 100, get
them from solving C 0 (x) = 0 with
C 0 (x) = 9 −
60000
x2
so
x2 = 60000/9 =
20000
3
p
p
so x = 100 2/3 ≈ 81 and y = 100 3/2 ≈ 122 (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)
x5 + 30x4 − 2x + 3 dx = x6 + 6x5 − x2 + 3x + C
6
Z
b)
cos(t) + 2 sin(t) dt = sin(t) − 2 cos(t) + C
Z
sin6 (2y)
+C
c)
cos(2y) sin5 (2y) dy =
12
Solution. I used the substitution u(y) = sin(2y) in c), so du = 2 cos(2y) dy.
Problem 3 Consider the function
f (t) = t165 .
a) What is the linearization L(t) of f (t) at t = 1?
b) Use differentials to approximate f (1.02). 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 (1.02) ≈ L(1.02) = 165 · 0.02 + 1 = dy + 1 = 3.3 + 1 = 4.3.
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 6
miles from the launch site. At a moment when the distance of the rocket from
the radar station is 8 miles, this distance increases by 2800 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 6 and one leg (vertical)
has length h, the height of the rocket. Then write equations between these
quantities:
z 2 = 62 + h2
and we know z 0 = 2800 at the moment when z = 8. Then we differentiate
the equation above:
2zz 0 = 2hh0
Substitute the numbers given for the particular moment in time. We have
to use z = 8 in the original equation to get
√
√
h = z 2 − 62 = 2 7
√
2 · 8 · 2800 = 4 · 7h0 .
Solving for h0 gives
(in miles per hour).
√
h0 = 1600 7 ≈ 4230