C Roettger, Spring 15
Math 165 - Exam 2D - solutions
Problem 1
A farmer wants to fence in a rectangle. She has a budget of $ 10,000, and
she wants to use sturdy fence, costing $ 10 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 biggest
area? 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 C = 10000. We also have the equations
C = 10x + 6y
A = xy
We use the first equation above with C = 10000 to eliminate one variable
from the second equation, eg x:
x=
10000 − 6y
10
so we can write C as a function of only one variable, namely y:
A(y) =
y(10000 − 6y)
10
There is no upper limit for y, but since y has to be the longer leg of the
triangle, we get x ≤ y and therefore
(10 + 6)y ≥ 10x + 6y = 10000
so y ≥ 1000/16 = 625. The domain for y is then (625, ∞). We proceed to
find critical points for in the domain. Besides the endpoint y = 625, get
them from solving A0 (y) = 0 with
A0 (y) =
10000 − 12y
10
so
10000
2500
=
≈ 830
12
3
(in feet). So x = (10000 − 6y)/10 = 500 (in feet). This y-value is indeed in
the domain. The sign pattern of A0 (y) on the domain is +− which means
that this last point really gives the maximum area.
y=
Problem 2 Find the following antiderivatives. Make sure to list the function
g amd its derivative if you are using Substitution.
Z
1
a)
x6 + 25x4 − 2x + 4 dx = x7 + 5x5 − x2 + 4x + C
7
Z
b)
3 sin(t) − cos(t) dt = −3 cos(t) − sin(t) + C
Z
sin5 (2y)
4
+C
c)
cos(2y) sin (2y) dy =
10
Solution. I used the substitution u(y) = sin(2y in c), so du = 2 cos(2y) dy.
Problem 3 Consider the function
f (t) = t2015 .
a) What is the linearization L(t) of f (t) at t = 1?
b) Use differentials to approximate f (1.003). Exact evaluations of the function f will not give credit.
Solution. a) We differentiate f ,
f 0 (t) = 2015t2014
so f 0 (1) = 2015. Also, f (1) = 1. Then the linearization is
L(t) = 2015(t − 1) + 1.
b) This just means using L(t) instead of f (t), which makes sort of sense
because 1.003 is close to 1 (although not very close, given the huge exponent/steep slope of the tangent line). We get
f (1.003) ≈ L(1.003) = 2015 · 0.003 + 1 = dy + 1 = 6, 045 + 1 = 7.045.
Problem 4 For the function
f (x) = (2x2 − 1)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. a) Since f 0 (x) is defined everywhere,
f 0 (x) = (4x − 4x2 + 2)e−2x = −2(2x2 − 2x − 1)e−2x .
we just need to solve f 0 (x) = 0. This is true exactly when
2x2 − 2x − 1 = 0
√
√
which has solutions x = (2 ± 12)/4 = (1 ± 3)/2.
0
b) The
√ of f (x) is − + −, so the function f (x) is increasing in
√ sign pattern
[(1 − 3)/2, (1 + 3)/2] and decreasing outside of this interval.√
c), d) From a) and b), f (x) √
has a local minimum at x = (1 − 3)/2 and a
local maximum at x = (1 + 3)/2.
Problem 5 A rocket is launched vertically and tracked by a radar station 3
miles from the launch site. At a moment when the distance of the rocket from
the radar station is 5 miles, this distance increases by 2700 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 = 32 + h2
and we know z 0 = 2700 at the moment when z = 5. Then we differentiate
the equation above:
2zz 0 = 2hh0
Substitute the numbers given for the particular moment in time. We have
to use z = 5 in the original equation to get
√
h = z 2 − 32 = 4
2 · 5 · 2700 = 8h0 .
Solving for h0 gives
h0 =
(in miles per hour).
27000
= 3375 ≈ 3380
8