C Roettger, Spring 15
Problem 1 A farmer wants to fence in a right triangle. She has a budget of $ 5000, and she wants to use sturdy fence, costing $ 10 per linear foot for the shorter leg, and cheap fence, costing $ 7 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 biggest area? 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 C = 5000. We also have the equations
C = 10 x + 7 y xy
A =
2
We use the first equation above with C = 5000 to eliminate one variable from the second equation, eg x : x =
5000 − 7 y
10 so we can write C as a function of only one variable, namely y :
A ( y ) = y (5000 − 7 y )
20
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 + 7) y ≥ 10 x + 7 y = 5000 so y ≥ 5000 / 17. The domain for y is then (5000 / 17 , ∞ ). We proceed to find critical points for in the domain. Besides the endpoint y = 5000 / 17, get them from solving A
0
( y ) = 0 with
A
0
( y ) =
5000 − 14 y
20

so y =
5000
14
=
2500
≈ 357
7
(in feet). So x = (5000 − 2500) / 10 = 250 (in feet). This y -value is indeed in the domain. The sign pattern of A
0
( y ) on the domain is + − which means that this last point really gives the maximum area.
Problem 2 Find the following antiderivatives. Make sure to list the function g and its derivative if you are using Substitution.
a) b)
Z x
4
+ 15 x
4
+ 2 x + 3 d x =
1
5 x
5
+ 3 x
5
+ x
2
+ 3 x + C
Z sin( t ) + 2 cos( t ) d t = − cos( t ) + 2 sin( t ) + C c)
Z cos(3 y ) sin
5
(3 y ) d y = sin
6
(3 y )
+ C
18
Solution.
I used the substitution u ( y ) = sin(3 y ) in c), so d u = 3 cos(3 y ) d y .
Problem 3 Consider the function f ( t ) = t
2015
.
a) What is the linearization L ( t ) of f ( t ) at t = 1?
b) Use differentials to approximate f (0 .
998). Exact evaluations of the function f will not give credit.
Solution.
a) We differentiate f , f
0
( t ) = 2015 t
2014 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 0.998 is close to 1 (although not very close, given the huge exponent/steep slope of the tangent line). We get f (0 .
998) ≈ L (0 .
998) = − 2015 · 0 .
002 + 1 = d y + 1 = − 4 .
03 + 1 = − 3 .
03 .

Problem 4 For the function f ( x ) = (2 x
2 − 1) e
− 2 x 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 ) = (4 x − 4 x
2
+ 2) e
− 2 x
= − 2(2 x
2 − 2 x − 1) e
− 2 x
.
we just need to solve f
0
( x ) = 0. This is true exactly when
2 x
2 − 2 x − 1 = 0 which has solutions x = (2 ±
√
12) / 4 = (1 ±
√
3) / 2.
b) The sign pattern of
[(1 − 3) / 2 , (1 + 3) c), d) From a) and b),
/ f ( f x
0
( x ) is − + − , so the function
) has a local minimum at x f
= (1
(
− x ) is increasing in
√
3) / 2 and a local maximum at x = (1+
√
3) / 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 9 miles, this distance increases by 3200 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
= 6
2
+ h
2 and we know z
0
= 2800 at the moment when z = 9. Then we differentiate the equation above:
2 zz
0
= 2 hh
0
Substitute the numbers given for the particular moment in time. We have to use z = 9 in the original equation to get h =
√ z 2 − 6 2 = 3
√
5
2 · 9 · 3200 = 6 ·
√
5 h
0
.
Solving for h
0 gives h
0
=
9600
√
5
≈ 4290
(in miles per hour).