Mock Midterm 2A
Note: The problems on this mock midterm have not necessarily been selected to allow them
to be easy to work without a calculator. The problems on the real midterm will not require
the use of a calculator.
(1) A manufacturer has been selling flashlights at $6 apiece, and at this price, consumers
have been buying 3000 flashlights per month. The manufacturer wishes to raise the
price and estimates that for each $1 increase in the price, 1000 fewer flashlights will
be sold each month. the manufacturer can produce the flashlights at a cost of $4 per
flashlight. At what price should the manufacturer sell the flashlights to generate the
greatest possible profit?
We want to maximize profit. First, let s be the number of dollars they raise the
price of the flashlights. Note that we can restrict s to 0 ≤ s ≤ 3, since if the
manufacturer raises the price more than $3, they will be selling a negative number
of flashlights, which makes no sense.Then we can set up the following equations.
Profit = Revenue − Cost
= (price charged)(number sold) − (cost to produce)(number produced)
= (6 + s)(3000 − 1000s) − (4)(3000 − 1000s)
= −1000s2 + 1000s + 6000
In order to find where profit is maximized, we need to take the derivative and set
it equal to zero. We then need to verify that the point that we find is indeed a
maximum rather than a minimum. We have
P 0 (s) = −2000s + 1000
0 = −2000s + 1000
1
2000s = 1000 ⇒ s =
2
We need to verify that this is indeed the desired maximum. Since there is only a
single critical point in our domain (indeed, on the entire real line), we can use the
second derivative test for absolute extrema. We have
P 00 (s) = −2000 < 0 for all s ∈ R
Hence s = 1/2 is indeed a maximum. This means that the manufacturer can maximize profit by selling the flashlights at $6.50 each. At that point, their profit would
be
P (1/2) = (6 + 1/2)(3000 − 1000 · 1/2) − (4)(3000 − 1000 · 1/2)
= 6250
(Note that you aren’t actually asked for the profit, so it would not be necessary to
give it on a test. It’s worth double checking though, because sometimes you will
be asked for the maximum profit rather than the price that should be charged to
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maximize profit. Also note that it is possible to do this problem by taking s to be
the price rather than the increase in price, but that it makes the algebra messier.)
(2) Find all points on the xy = 16y 2 + x where the tangent line is horizontal.
The tangent line will be horizontal precisely at those points where the derivative is
zero. Hence we begin the problem by finding the derivative. Since we cannot easily
separate the x and y in this function, we will use implicit differentiation and then
dy
. We have
solve for
dx
dy
dy
1·y+x·
= 32y ·
+1
dx
dx
dy
dy
x·
− 32y ·
=1−1·y
dx
dx
dy
(x − 32y)
=1−y
dx
dy
1−y
=
dx
x − 32y
To see where this is zero, we observe that a rational function can only be zero when
the numerator is zero. So it suffices to set the numerator to zero. We have
1−y =0⇒y =1
Now we know the y coordinate of any points where the tangent line is horizontal.
We still need to find the x coordinates. In order to do this, we’ll plug y = 1 into the
original function and solve for x. We have
x · 1 = 16 · 12 + x
x = 16 + x
0 = 16
Since it is clear that that 16 6= 0, we have no points where the tangent line is
horizontal.
(3) Given h(x) = 3x4 − 4x2 + 3,
(a) Identify any asymptotes for h(x).
(b) Find and classify all critical points of h(x).
(c) Find the inflection points and intervals of concavity for h(x).
(d) Sketch a graph of h(x). Be sure to include all important features of the graph.
(a) It should be clear that h(x) does not have any vertical asymptotes because it is
defined everywhere. To check whether or not there are horizontal asymptotes,
we take
lim 3x4 − 4x2 + 3 = +∞
x→+∞
and
lim 3x4 − 4x2 + 3 = +∞
x→−∞
Since the function is going to infinity rather than to a finite number as x → ±∞,
we have no horizontal asymptotes either.
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(b) To find critical points, we take the derivative and set it to zero and solve. We
have
h0 (x) = 12x3 − 8x = 0
4x(3x2 − 2) = 0
p
p
4x(x − 2/3)(x + 2/3) = 0
p
p
So our critical numbers are x = −p 2/3, x = 0,
and
x
=
2/3, which divide
p
p the
real line into
p the intervals x < − 2/3 and − 2/3 < x < 0 and 0 < x < 2/3
and x > 2/3. Now, we need to choose numbers in each of these
p intervals to
see where the function is increasing and decreasing. Note that 2/3 ≈ .816.
We choose the numbers −1, −1/2, 1/2, and 1, respectively, from these intervals.
We obtain
h0 (−1) = (−)(−)(−) = (−)
h0 (−1/2) = (−)(−)(+) = (+)
h0 (1/2) = (+)(−)(+) = (−)
h0 (1) = (+)(+)(+) = (+)
p
p
p
p
p
So
h(x)
has
relative
minima
at
(−
2/3,
f
(−
2/3))
=
(−
2/3,
5/3)
and
(
2/3,
f
(
2/3)) =
p
( 2/3, 5/3) and a relative maximum at (0, f (0)) = (0, 3)
(c) To find intervals of concavity we need to compute the second derivative. We
have
h00 (x) = 36x2 − 8
We set this to 0 to obtain
p
0 = 36x2 − 8 ⇒ 36x2 = 8 ⇒ x = ± 2/9
p
p
which
divides
our
real
line
into
the
intervals
x
<
−
2/9
and
−
2/9 < x <
p
p
2/9 and x > 2/9. Choosing the numbers -1, 0, and 1, respectively from
these intervals, we have
h00 (−1) = 36 − 8 > 0
h00 (0) = 0 − 8 < 0
h00 (1) = 36 − 8 > 0
p
p
So our function is concave down on −
x <
2/9 and concave up
p 2/9 < p
elsewhere. It has inflection points at − 2/9 and 2/9.
(d) Our graph looks like this:
3
6
5
4
3
2
1
-4
-3
-2
-1
0
1
2
3
4
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(4) Find an equation of the line that is tangent to the graph of f (x) = (x2 − 3)5 (2x − 1)3
at x = 2.
To find the equation of the tangent line, we first need to find the slope of the
tangent line, and in order to do that, we need to find the derivative of the function
and plug in our value for x. We have
f 0 (x) = 5(x2 − 3)4 (2x)(2x − 1)3 + (x2 − 3)5 · 3(2x − 1)2 (2)
= 5(x2 − 3)4 (2x)(2x − 1)3 + (x2 − 3)5 · 3(2x − 1)2 (2)
= (x2 − 3)4 (2x − 1)2 [10x · (2x − 1) + 6 · (x2 − 3)]
= (x2 − 3)4 (2x − 1)2 [20x2 − 10x + 6x2 − 18]
= (x2 − 3)4 (2x − 1)2 (26x2 − 10x − 18)
f 0 (2) = (4 − 3)4 (4 − 1)2 (26 · 4 − 10 · 2 − 18)
= 1 · 9 · 66 = 594
We also need to know y = f (2). We have
f (2) = (22 − 3)5 (2 · 2 − 1)3 = 15 · 33 = 27
So our tangent line is given by
y − 27 = 594(x − 2) ⇒ y = 594x − 1161.
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(5) An art gallery offers 50 prints by a famous artist. If each print in the limited edition
is priced at p dollars, it is expected that q = 500 − 2p prints will be sold.
(a) What limitations are there on the possible range of the price p?
(b) Find the elasticity of demand. Determine the values of p for which the demand
is elastic, inelastic, and of unit elasticity.
(c) Interpret the results of part (b) in terms of the behavior of the total revenue as
a function of unit price p?
(d) If you were the owner of the gallery, what price would you charge for each print?
Explain the reasoning behind your decision.
(a) Since there are only 50 prints available, the least that the prints can be sold for
solves
500 − 2pmin = 50 ⇒ 450 = 2pmin ⇒ pmin = 225
Since they cannot sell a negative number of prints, the most that the prints can
be sold for solves
500 − 2pmax = 0 ⇒ 500 = 2pmax ⇒ pmax = 250
So the prints must be priced between $225 and $250.
(b) To compute the elasticity of demand, we have
p dq
q dp
p
=−
(−2)
500 − 2p
2p
=
500 − 2p
E(p) = −
To see where E(p) is of unit elasticity, we take
1=
2p
⇒ 500 − 2p = 2p ⇒ 500 = 4p ⇒ 125 = p
500 − 2p
so we have unit elasticity when we charge $125. Observe that this is outside of
the range of the prices we can charge. If we choose p = 100 < 125, we have
200
E(100) =
<1
300
and so demand is inelastic when the price is lower than $125. If we chose
p = 200 > 125, we have
400
E(200) =
>1
100
and so demand is elastic when the price is higher than $125.
(c) If we had an unlimited number of prints available for sale, our revenue would
be maximized when we priced the prints at $125. However, we don’t have an
unlimited number of prints available, and at $125, the demand for prints would
greatly outstrip the supply.
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(d) I would price the prints at $225 because it maximizes the revenue function,
R(p) = 500p − 2p2 on the interval on which we can actually price the prints.
Since
R0 (p) = 500 − 4p = 0 ⇒ p = 125
is the only critical point, and it is not in our interval, the maximum must occur
at one of our two endpoints. We have R(225) = 500 · 225 − 2 · 2252 = 11250 and
R(250) = 500 · 250 − 2 · 2502 = 0.
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