# Math 165 - Practice Exam 2 - solutions

```C Roettger, Spring 15
Math 165 - Practice Exam 2 - solutions
Problem 1 Suppose a balloon in the shape of a cylinder is inflated, so that
its volume increases at a rate of 20π cubic centimeters per second. At a
time when its radius is 6 cm and its height 4 cm, the height of the cylinder
increases at a rate of 0.2 cm per second. What is the rate of change of the
radius at this moment in time? Round the answer to four digits after the
decimal point.
Solution. First, we introduce names for the relevant quantities. Let V, h, r
the volume, height, and radius of the cylinder, respectively. Then write
equations between these quantities:
V = πhr2
and we know V 0 = 20π (at all times) and at the time when r = 6 and h = 4,
h0 = 0.2. Then we differentiate the equation above:
V 0 = π(r2 h0 + 2hrr0 ).
Substitute the numbers given for the particular moment in time:
20π = π(36 &middot; 0.2 + 48r0 ).
Solving for r0 gives
r0 =
4
20 − 7.2
=
≈ 0.2667
48
15
(in cm per second).
Problem 2 Suppose an asteroid is observed to have a current speed of
v0 = 1, 200
km
s
Simulations of the asteroid’s projected path lead to this estimate of the distance d of the asteroid to Earth at the time when it will be closest,
d = 180, 000 + 75v0 − 0.001v03
1
The asteroid’s speed can only be measured with limited accuracy. But the
astronomers are sure that
v0 = 1, 200 &plusmn; 40
km
s
Use differentials to estimate the corresponding change in the value of d. Exact
evaluations of the function d will not give credit.
Solution. We differentiate d (with regard to the input parameter v0 !) and
substitute v0 = 1200, so for this value of v0
d0 = 75 − 0.003v02 = 4245.
So the absolute change in d is estimated to be
|∆d| ≈ 4245 &middot; 40 = 169, 800
Compared to d(1200) = 162, 000, the relative change is
169, 800
≈ 104%
162, 000
This means the change could be so big that the true value of d was negative!
While this does not make sense physically, it surely would be bad news for
our planet.
Fortunately, a repeat of the measurement gave v0 = 1190.
Problem 3 Suppose you know that the derivative of an unknown function
is
(x + 3)2 (x + 1)(x − 2)(x − 5)
f 0 (x) =
x2 + 1
Find all critical points of f (x). Use the First Derivative Test to determine
which are relative maxima and which are minima.
Solution. The critical points are x = −3, −1, 2, 5 because these are the
only x-values that give f 0 (x) = 0.
The sign pattern of f 0 (x) is − − + − + with changes at the critical points
x = −1, 2, 5 but no sign change at −3. So x = −1, 5 are relative minima,
x = 2 is a relative maximum. x = −3 is not a relative extremum.
Problem 4 Find all relative extrema of
f (x) = x4 (x2 − 2)
2
Use the Second Derivative Test to determine whether they are relative minima or maxima.
Solution. First, we find the critical points: from
f 0 (x) = 4x3 (x2 − 2) + 2x5 = 2x3 (3x2 − 4)
√
we can see that they are x = 0, &plusmn;2 3/3. Then find the second derivative,
f 00 (x) = 30x4 − 24x2
√
and substitute the critical values of x. For x = &plusmn;2 3/3, we find f 00 (x) =
4(40 − 24)/3 &gt; 0, so these are relative minima. For x = 0, the test is
inconclusive. Using the SDT, we would just stop here. If we switch to using
the FDT, we see that f 0 (x) changes sign from positive to negative at x = 0,
so x = 0 is a local maximum. Not part of this problem: compare to a
plot of f (x) and see how our results match up with the plot.
Problem 5 Find all relative extrema of
f (x) =
x2 − 2
x+3
and tell which ones are relative minima and which ones are relative maxima.
Identify the intervals on which f (x) is increasing and the intervals on which
f (x) is decreasing.
Solution. The only singular point is x = −3, f (x) is not defined there. For
stationary points, we find
f 0 (x) =
2x(x + 3) − (x2 − 2)
x2 + 6x + 2
=
(x + 3)2
(x + 3)2
3
√
with zeros at x = −3 &plusmn; 7 (find these by solving x2 + 6x + 2 = 0). The sign
pattern of f 0 (x) is + − || − + (the double bars stand for a singular point). In
more detail,
√
√
−3 − 7
−3
−3 + 7
sign of f’ +
−
|| −
+
√
√
Therefore −3− 7 is a relative maximum, and −3+ 7 is a relative
√ minimum.
7) and (−3
The
function
f
(x)
is
increasing
on
the
intervals
(−∞,
−3
−
√
√
√+
7, ∞). It is decreasing on the intervals (−3 − 7, −3) and (−3, −3 + 7).
Problem 6 Find two positive numbers x, y such that xy = 1 and
1 2
+
x y
is minimal.
Solution. First, express the quantity 1/x + 2/y as function of one variable
alone:
1
f (x) = + 2x.
x
Next, minimize f (x). A singular point is obviously x = 0, for stationary
points we look at
1
f 0 (x) = − 2 + 2 = 0
x
√
√
1
which gives x = &plusmn; 2 2. Discarding the negative value, x = 12 2 must give
the (only) minimum. We can confirm this using the FDT or SDT.
Problem 7 Suppose the cost of making x units of FanCytm hairdryers is
C(x) = 1500 + 60x + 0.002x3
(dollars). If you can sell all of them at a price of 75 dollars each, find the
number x which will give you maximum profit. Round the answer to the
4
nearest whole integer.
Solution. The revenue from selling x units of hairdryers is 75x, so the profit
from x units is (revenue minus cost)
P (x) = −0.002x3 + 15x − 1500
The domain for x is [0, ∞) (there would be practical limitations on how big
x could be, but we don’t know about these). Then differentiate,
P 0 (x) = −0.006x2 + 15
which is zero for x2 = &plusmn;15/0.006 = 2500, so x = 50 (no rounding necessary,
actually). This is indeed a maximum, because P 0 (x) changes sign from + to
− there.
5
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