Download Midterm Exam 2 Solutions

Midterm Exam 2
Math 0120
Spring 2015
Solutions
100 points total
1.
For the function f (x) =
1
determine the x-coordinates of all inflection points of f .
2x − 6
Then find the open intervals on which f is concave up and down.
Solution:
f (x) = (2x − 6)−1 , f 0 (x) = −2(2x − 6)−2 , f 00 (x) = 8(2x − 6)−3 =
8
.
(2x − 6)3
f 00 is never zero. It is undefined at x = 3.
Hence IP is at x = 3.
[To be more precise there is no IP on the curve since 3 is not in the domain and at x = 3 the
curve changes its concavity. If you answer is: IP is at x = 3, then I count it as right.]
f 00 (x) < 0 on (−∞, 3)
f 00 (x) > 0 on (3, ∞)
Correspondingly, f (x) is concave down on (−∞, 3),
f (x) is concave up on (3, ∞)
2. The manager of a large apartment complex knows from experience that 120 units will be occupied
if the rent is 600 dollars per month. A market survey suggests that, on the average, one additional
unit will remain vacant for each 6 dollar increase in rent. Similarly, one additional unit will be
occupied for each 6 dollar decrease in rent. What rent should the manager charge to maximize
revenue?
Solution:
Let x be the number of price reductions (x can be negative if the price is increased).
Then the price is p(x) = 600 − 6x, the quantity of rented appartments is q(x) = 120 + x, and
the revenue is R(x) = p(x)q(x) = 6(100 − x)(120 + x). To find maximum of R(x) we apply the
optimization process when −120 ≤ x ≤ 100.
R0 (x) = 6[−(120 + x) + (100 − x)] = 6[−20 − 2x] = 0, x = −10. R0 (x) is defined everywhere.
The only CN is x = −10. R00 (x) = 6(−2) = −12 < 0 and the graph of R(x) is concave down on
the entire number line. Hence R(x) attains its absolute maximum at x = −10.
The rent that the manager should charge to maximize revenue is p(−10) = 600 + 60 = 660 dollars.
3.
Find the slope of the tangent to the curve 2x3 + 3xy + y 2 = 12 at (1, 2).
Solution:
Implicit differentiation gives 6x2 + 3y + 3xy 0 + 2yy 0 = 0.
At (1, 2) we have 6 + 6 + 3y 0 (1) + 4y 0 (1) = 0, 7y 0 (1) = −12, y 0 (1) = − 12
7 .
The slope is m = − 12
7
4.
Two people on bikes start riding from the same point and at the same time. Person A starts
riding west at a rate of 3 m/sec and Person B starts riding south at 4 m/sec. At what rate is the
distance separating the two people changing 2 seconds after they started riding?
Solution:
Let x be a distance between a position of the person A and the initial point, y be a
distance between a position of the person B and the initial point, and s be a distance between
two persons. Their paths form a right triangle. We have x2 + y 2 = s2 .
Differentiation of the equation with respect to t gives
2xx0 + 2yy 0 = 2ss0 or xx0 + yy 0 = ss0 where x0 = 3 and y 0 = 4.
2 seconds after they started riding x = 3 · 2 = 6, y = 4 · 2 = 8, s =
So, xx0 + yy 0 = ss0
√
⇔ 6 · 3 + 8 · 4 = 10s0 ⇔ 50 = 10s0 ⇔ s0 =
62 + 82 =
ds
dt
√
100 = 10
= 5 m/sec.
5. In a certain year, a total of 6 million passengers took a cruise vacation. The global cruise industry
is growing at 8% per year. In how many years the number of passegers will double. Leave your
answer in exact form.
Solution:
8% = 0.08. Let P0 = 6 millions.
P (t) = P0 (1 + 0.08)t = P0 (1.08)t = 2P0 , (1.08)t = 2, ln(1.08)t = ln 2, t ln 1.08 = ln 2
t=
6.
ln 2
years.
ln 1.08
Find the relative growth rate for each of the following functions.
(a)
f (x) = 5e3x − x3
Solution:
f 0 (x) = 15e3x − 3x2 ,
f 0 (x)
15e3x − 3x2
=
f (x)
5e3x − x3
(b) h(t) = 4t ln t
Solution:
h0 (t) = 4t ln 4 ln t + 4t · 1t ,
4t ln 4 ln t + 4t ·
h0 (t)
=
h(t)
4t ln t
1
t
=
ln 4 ln t +
ln t
1
t
= ln 4 +
Page 2
1
t ln t
7. Demand for movie tickets to a certain theater depends on the ticket price p, in dollars, according
√
to the demand function D(p) = 200 15 − p
Ticket price is currently set at 6 dollars.
(a)
Find the elasticity of demand E(p) at the current ticket price.
Solution:
E(6) =
(b)
100
= −√
,
15 − p
p · √100
−p · D0 (p)
p
15−p
√
E(p) =
=
=
,
D(p)
2(15 − p)
200 15 − p
6
1
6
=
= .
2(15 − 6)
18
3
Is the demand elastic at the current ticket price? Should the price be lowered or raised?
Solution:
(c)
D0 (p)
The demand is inelastic because E(6) =
1
3
< 1. The price should be raised.
Find the ticket price which will maximize revenue.
Solution:
E(p) = 1,
p
= 1, p = 30 − 2p, 3p = 30,
2(15 − p)
p = 10 dollars.
bonus problem. [5 points extra] Find y 00 (1) if x2 y + y 3 = 2
Solution:
If x = 1 then 12 · y(1) + [y(1)]3 = 2. Hence y(1) = 1.
We differentiate the equation twice to get y 00 :
(1) 2xy + x2 y 0 + 3y 2 y 0 = 0. When x = 1, y = 1 we have 2 + y 0 + 3y 0 = 0 and y 0 (1) = − 42 = − 12
(2) 2y + 2xy 0 + 2xy 0 + x2 y 00 + 6y(y 0 )2 + 3y 2 y 00 = 0.
When x = 1, y = 1, and y 0 = − 21 we have 2 − 1 − 1 + y 00 +
Hence y 00 (1) = − 38
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6
4
+ 3y 00 = 0 or 4y 00 = − 32