math 115 — practice for exam 2 - university of …mconger/dhsp/115packet2soln.pdfhence x= 0.8,5...

Math 115 — Practice for Exam 2

Generated November 12, 2017

Name: SOLUTIONS

Instructor: Section Number:

1. This exam has 18 questions. Note that the problems are not of equal difficulty, so you may want toskip over and return to a problem on which you are stuck.

2. Do not separate the pages of the exam. If any pages do become separated, write your name on themand point them out to your instructor when you hand in the exam.

3. Please read the instructions for each individual exercise carefully. One of the skills being tested onthis exam is your ability to interpret questions, so instructors will not answer questions about examproblems during the exam.

4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that thegraders can see not only the answer but also how you obtained it. Include units in your answers whereappropriate.

5. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).However, you must show work for any calculation which we have learned how to do in this course. Youare also allowed two sides of a 3′′ × 5′′ note card.

6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of thegraph, and to write out the entries of the table that you use.

7. You must use the methods learned in this course to solve all problems.

Semester Exam Problem Name Points Score

Winter 2017 2 5 12

Winter 2014 3 3 potential energy 11

Fall 2009 2 4 UM mug 12

Winter 2017 2 6 corn snake 9

Fall 2009 2 6 gravity 14

Fall 2016 2 11 6

Winter 2012 2 6 16

Winter 2011 2 4 antihistamine 15

Fall 2016 2 6 14

Fall 2009 2 7 Kampyle of Eudoxus 14

Winter 2014 2 11 5

Winter 2014 2 4 ship 8

Fall 2014 2 9 caffeine 10

Fall 2014 2 4 garden 12

Winter 2007 3 4 octopus2 12

Fall 2002 3 10 trough 10

Winter 2016 2 6 11

Winter 2013 2 6 9

Total 200

Recommended time (based on points): 190 minutes

Math 115 / Exam 2 (March 22, 2017) page 6

5. [12 points] Let

f(x) = x(x− 4)4/5e−x and f ′(x) =(5− x)(5x− 4)e−x

5 5√x− 4

.

Note that the domain of f(x) is (−∞,∞).

a. [6 points] Find all values of x at which f(x) has a local extremum. Use calculus to findand justify your answers, and be sure to show enough evidence to demonstrate that youhave found all local extrema. For each answer blank below, write none if appropriate.

Solution: First we look for:

• Values of x for which f ′(x) = 0:

(5− x)(5x− 4)e−x

5 5√x− 4

= 0

(5− x)(5x− 4)e−x = 0

5− x = 0 or 5x− 4 = 0

x = 5 or x = 0.8.

Notice that e−x 6= 0 for all values o x. Hence the only solutions are x = 0.8 and x = 5.

• Values of x for which f ′(x) is undefined: The function f ′(x) = (5−x)(5x−4)e−x

5 5√x−4

is

undefined when.

5 5√x− 4 = 0

5√x− 4 = 0

x− 4 = 0 which yields x = 4.

Finally we notice that all of these points are in the domain of f(x), hence the criticalpoints of f(x) are x = 0.8, 4, 5.

We use the first derivative test to classify all the critical points of f(x) by either findingthe signs of f ′(x) around the critical points or computing its values:

f ′(0) =(+)(−)(+)

− = + or f ′(0) ≈ 3.031

f ′(1) =(+)(+)(+)

− = − or f ′(1) ≈ −0.236

f ′(4.5) =(+)(+)(+)

+= + or f ′(4.5) ≈ 0.023

f ′(6) =(−)(+)(+)

+= − or f ′(6) ≈ −0.011

Hence x = 0.8, 5 are local maximums and x = 4 is a local minimum.

Answer:

Local max(es) at x = 0.8, 5 Local min(s) at x = 4

University of Michigan Department of Mathematics Winter, 2017 Math 115 Exam 2 Problem 5 Solution


b. [6 points] Find the values of x for which f(x) attains a global maximum and global mini-mum. Use calculus to find and justify your answers, and be sure to show enough evidenceto demonstrate that you have found all global extrema. Write none if appropriate.

Solution:

x 0.8 4 5

f(x) 0.911 0 0.033and

limx→∞

x(x− 4)4/5e−x = limx→∞

x(x4/5)

ex= lim

x→∞

x9/5

ex= 0

limx→−∞

x(x− 4)4/5e−x = limx→−∞

x9/5e−x = −∞.

The last limit follows from the fact that limx→−∞

x9/5 = −∞ and limx→−∞

e−x = ∞. Hence

the function f(x) has a global maximum at x = 0.8 and no global minimum.


Math 115 / Final (April 28, 2014) page 4

3. [11 points] For positive constants a and b, the potential energy of a particle is given by

U(x) = a

(

5b2

x2−

3b

x

)

.

Assume that the domain of U(x) is the interval (0,∞).

a. [2 points] Find the asymptotes of U(x). If there are none of a particular type, write none.

Solution: We can get a common denominator and write

U(x) = a5b2 − 3bx

x2.

We see that there is a vertical asymptote at x = 0, where the denominator is zero, and ahorizontal asymptote at U = 0, since the degree of the denominator is greater than the degree ofthe numerator.

Answer: Vertical asymptote(s): x = 0 Horizontal asymptote(s): U = 0

b. [6 points] Find the x-coordinates of all local maxima and minima of U(x) in the domain(0,∞). If there are none of a particular type, write none. You must use calculus to findand justify your answers. Be sure to provide enough evidence to justify your answers fully.

Solution: First we find critical points by looking at where U ′(x) is undefined or zero. We have

U ′(x) = a

(

−

10b2

x3+

3b

x2

)

= a3bx− 10b2

x3.

There are no points in the domain of U(x) where U ′(x) is undefined, but U ′(x) has a zero where3bx− 10b2 = 0, or x = 10b

3.

To classify this critical point, we can use the First or Second Derivative Test. We will use theSecond Derivative Test here, so we compute

U ′′(x) = a

(

30b2

x4−

6b

x3

)

=6ab

x4(5b− x).

Since a, b, and x4 are always positive and 5b−x is positive at x = 10b

3, we see that U ′′

(

10b

3

)

> 0,

and hence x = 10b

3is a local minimum.

There are no other critical poitns to consider, so there are no local maxima.

Answer: Local max(es) at x = none Local min(s) at x =

10b

3

c. [3 points] Suppose U(x) has an inflection point at (6,−14). Find the values of a and b.Show your work, but you do not need to verify that this point is an inflection point.

Solution: We already found U ′′(x) = 6ab

x4 (5b − x), so we see that the only potential inflectionpoint occurs at x = 5b, the only place in the domain of U(x) where U ′′(x) is zero or undefined.Hence 5b = 6 or b = 1.2.Plugging in x = 6, b = 1.2, and U = −14 into the original equation for U(x) yields

−14 = a

(

1

5−

3

5

)

= −

2a

5

and hence a = 35.

Answer: a = 35 and b = 1.2University of Michigan Department of Mathematics Winter, 2014 Math 115 Exam 3 Problem 3 (potential energy) Solution

Math 115 / Exam 2 (November 24, 2009) page 6

4. [12 points] In preparation for the holidays, a local bookstore is planning to sell mugs of avariety of shapes. Suppose that the amount of liquid in a “UM” mug if filled to a depth of hcm is L(h) = Uh(3M2

− 3Mh + h2) cm3 for U,M > 0.

a. [4 points] Find and classify any critical points of L on the interval (0, 5M).

Solution: Taking the derivative gives

L′(h) = U(3M2− 6Mh + 3h2) = 3U(M2

− 2Mh + h2) = 3U(M − h)2.

Thus, the only critical point occurs at h = M . Note that the factor (M − h)2 is positivefor all h, so the function is increasing to the left of h = M and to the right of h = M .Thus, the critical point is neither a local maximum nor a local minimum.

b. [2 points] Determine any points of inflection of L on the interval (0, 5M).

Solution: The second derivative, L′′(h) = −6U(M − h), shows a potential inflectionpoint at h = M . The sign of the factor −6U is always negative. The sign of the factor(M − h) is positive to the left of h = M and negative to the right. Thus, the productgives us L′′(h) < 0 for h < M , and L′′(h) > 0 for h > M , and the function changes fromconcave down to concave up at h = M , so L has an inflection point at h = M .

c. [6 points] Suppose you are pouring coffee into a “UM” mug at a rate of 15 cm3 per second.At what rate is the depth of the coffee in the mug changing when the coffee reaches adepth of 4 cm in the mug?

Solution: Given dL/dt = 15 cm3/s, we want to find dh/dt when h = 4 cm. We know

dL

dt=

dL

dh·

dh

dt,

so, when h = 4, we have

15 = 3U(M − 4)2 ·dh

dt

anddh

dt=

15

3U(M − 4)2cm/second.

University of Michigan Department of Mathematics Fall, 2009 Math 115 Exam 2 Problem 4 (UM mug) Solution


6. [9 points] A group of biology students is studying the length L of a newborn corn snake (incm) as a function of its weight w (in grams). That is, L = G(w). A table of values of G(w) isshown below.

w 5 10 15 20 25

G(w) 24.5 31.6 38.7 44.7 50

G′(w) 2.23 1.58 1.30 1.12 1.05

Assume that G′(w) is a differentiable and decreasing function for 0 < w < 25.

a. [2 points] Find a formula for H(w), the tangent line approximation of G(w) near w = 20.

Solution: The formula for is H(w) = G(20) + G′(20)(w − 20). From the table we getH(w) = 44.7 + 1.12(w − 20).

b. [1 point] Use the tangent line approximation of G(w) near w = 20 to approximate thelength of a corn snake that weighs 22 grams.

Solution: G(22) ≈ H(22) = 1.12(22− 20) + 44.7 = 46.94 cm.

c. [2 points] Is your answer in part (b) an overestimate or an underestimate? Circle youranswer and write a sentence to justify it.

Solution:

Circle one: Overestimate Underestimate cannot be determined

Justification:

Since G′(w) is a differentiable and decreasing function for 0 < w < 25, then G(w) isconcave down on 0 < w < 25. Hence the values of the tangent line approximation H(w)will be larger than the actual values of G(w) for 0 < w < 25.

d. [4 points] In their study of the growth of corn snakes, they found the results of a recentarticle that states that the average weight w of a corn snake (in grams) t weeks after beingborn is given by w = 1

5t2. Let S(t) = G

(

1

5t2)

be the length of a corn snake t weeks af-ter being born. Find a formula for P (t), the tangent line approximation of S(t) near t = 5.

Solution: The formula for the tangent line approximation P (t) isP (t) = S(5) + S′(5)(t − 5). Since S(t) = G(1

5t2), then S′(t) = 2

5t · G′(1

5t2). Using these

formulas we get that S(5) = G(15(52)) = G(5) = 25.4 and S′(5) = 2 ·G′(5) = 4.46.

Answer: P (t) = 24.5+4.46(t− 5) = 4.46t+2.2



6. [14 points] The force F due to gravity on a body at height h above the surface of the earth isgiven by

F (h) =mgR2

(R + h)2

where m is the mass of the body, g is the acceleration due to gravity at sea level (g < 0), andR is the radius of the earth.

a. [3 points] Compute F ′(h).

Solution:

F ′(h) =−2mgR2

(R + h)3

b. [3 points] Compute F ′′(h).

Solution:

F ′′(h) =6mgR2

(R + h)4

c. [5 points] Find the best linear approximation to F at h = 0.

Solution: Since F (0) = mg and F ′(0) = −2mgR, the best linear approximation to F ath = 0 is given by the equation

L(h) = mg −

2mg

R· h.

Hence, near h = 0, the linear approximation gives

F (h) ≈ mg −

2mg

R· h.

d. [3 points] Does your approximation from part (c) give an overestimate or an underestimateof F? Why?

Solution: Since F ′′ is negative for all h (due to g < 0), the function is concave down, sothe tangent lies above the curve and the estimate is an overestimate.

University of Michigan Department of Mathematics Fall, 2009 Math 115 Exam 2 Problem 6 (gravity) Solution


10. [4 points] Let a and b be constants. Consider the curve C defined by the equation

cos(ax) + by ln(x) = 3 + y3.

Find a formula for dydx

in terms of x and y. The constants a and b may appear in your answer.To earn credit for this problem, you must compute this by hand and show every step of yourwork clearly.

Solution: We use implicit differentiation.

d

dx(cos(ax) + by ln(x)) =

d

dx(3 + y3)

−a sin(ax) +by

x+ b ln(x)

dy

dx= 3y2

dy

dx

dy

dx=

byx− a sin(ax)

3y2 − b ln(x)

Answer:dy

dx=

byx− a sin(ax)

3y2 − b ln(x)

11. [6 points] Let h(x) = xx. For this problem, it may be helpful to know the following formulas:

h′(x) = xx (ln(x) + 1) and h′′(x) = xx(

1

x+ (ln(x) + 1)2

)

.

a. [2 points] Write a formula for p(x), the local linearization of h(x) near x = 1.

Solution: h(1) = 1 and h′(1) = 11(ln(1) + 1) = 1, so p(x) = 1 + 1 · (x− 1) = x.

Answer: p(x) = x

b. [4 points] Write a formula for u(x), the quadratic approximation of h(x) at x = 1. (Recallthat a formula for the quadratic approximation Q(x) of a function f(x) at x = a is

Q(x) = f(a) + f ′(a)(x− a) + f ′′(a)2 (x− a)2.)

Solution: h′′(1) = 1(1 + (0 + 1)2) = 2, so u(x) = 1 + (x− 1) + 22(x− 1)2 = x2 − x− 1.

Answer: u(x) = 1 + (x− 1) + (x− 1)2 (= x2 − x− 1)

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 2 Problem 11 Solution


6. [16 points] Consider the piecewise linear function f(x) graphed below:

8

10

f(x)

x

(10,−6)

(4, 6)

For each function g(x), find the value of g′(3):

a. [4 points] g(x) = sin(

[f(x)]3)

Solution:

g′(x) = cos(f(x)3) · 3f(x)2 · f ′(x)

g′(3) = cos(73) · 3 · 72 · (−1) = 124.0442.

b. [4 points] g(x) =f(x2)

x

Solution:

g′(x) =x · f ′(x2) · 2x− f(x2)

x2

g′(3) =3(−3)6 − (−3)

9= −5.667.

c. [4 points] g(x) = ln(f(x)) + f(2)

Solution:

g′(x) =1

f(x)f ′(x) + 0

g′(3) =1

7· (−1) =

−1

7.

d. [4 points] g(x) = f−1(x)

Solution:

g′(x) =1

f ′(f−1(x))

g′(3) =1

f ′(6)= −

2

3.



4. [15 points] A model for the amount of an antihistamine in the bloodstream after a patienttakes a dose of the drug gives the amount, a, as a function of time, t, to be a(t) = A(e−t−e−kt).In this equation, A is a measure of the dose of antihistamine given to the patient, and k isa transfer rate between the gastrointestinal tract and the bloodstream. A and k are positiveconstants, and for pharmaceuticals like antihistamine, k > 1.

a. [5 points] Find the location t = Tm of the non-zero critical point of a(t).

Solution: The maximum will occur at an endpoint or at a critical point, when a′(t) = 0.The critical points are thus where a′(t) = A(−e−t + ke−kt) = 0. Solving, we havee−t = ke−kt, so that e(k−1)t = k, or t = Tm = 1

k−1 ln(k).

b. [3 points] Explain why t = Tm is a global maximum of a(t) by referring to the expressionfor a(t) or a′(t).

Solution: Note that a′(0) = A(k−1) > 0 and that for large t, a′(t) = A(−e−t+ke−kt) < 0(k > 1 guarantees that the second exponential decays much faster than the first). Thusthe critical point must be a maximum. In addition, because t = Tm this is the onlycritical point we know it must be the global maximum.Alternately, note that a(0) = 0. Because k > 0, a(t) ≥ 0 for all t (the exponentialinvolving −kt will decay faster than e−t). And for large t, a(t) → 0. Thus at t = Tm,a(t) must take on a maximum value, and because it is the only critical point this mustbe the global maximum.

c. [4 points] The function a(t) has a single inflection point. Find the location t = TI of thisinflection point. You do not need to prove that this is an inflection point.

Solution: To find inflection points, we look for where a′′(t) = 0. This gives a′′(t) =A(e−t − k2e−kt) = 0. Solving for t, we have (similarly to in (a)) e(k−1)t = k2, so thatt = TI =

1k−1 ln(k2) = 2

k−1 ln(k).(We could show that this is an inflection point by a similar argument to (b): becausek > 1 we know that a′′(0) < 0 and as t increases a′′(t) must eventually become positive.Thus this is an inflection point.)

d. [3 points] Using your expression for Tm from (a), find the rate at which Tm changes as kchanges.

Solution: We have Tm = 1k−1 ln(k). Thus

dTm

dk= −

1

(k − 1)2ln(k) +

1

k(k − 1).

University of Michigan Department of Mathematics Winter, 2011 Math 115 Exam 2 Problem 4 (antihistamine) Solution


6. [14 points] The entire graph of a function g(x) is shown below. Note that the graph of g(x)has a horizontal tangent line at x = 1 and a sharp corner at x = 4.

1 2 3 4 5

1

2

3

4

5y = g(x)

x

y

For each of the questions below, circle all of the available correct answers.(Circle none of these if none of the available choices are correct.)

a. [2 points] At which of the following values of x does g(x) appear to have a critical point?

x = 1 x = 2 x = 3 x = 4 none of these

b. [2 points] At which of the following values of x does g(x) attain a local maximum?

x = 1 x = 2 x = 3 x = 4 none of these

c. [6 points] Let L(x) be the local linearization of g(x) near x = 3. Circle all of the statementsthat are true.

L(3) > g(3)

L(3) = g(3)

L(3) < g(3)

L′(3) > g′(3)

L′(3) = g′(3)

L′(3) < g′(3)

L(2.5) > g(2.5)

L(2.5) = g(2.5)

L(2.5) < g(2.5)

L′(2.5) > g′(2.5)

L′(2.5) = g′(2.5)

L′(2.5) < g′(2.5)

L(0) > g(0)

L(0) = g(0)

L(0) < g(0)

L(5) > g(5)

L(5) = g(5)

L(5) < g(5)

none of these

d. [2 points] On which of the following intervals does g(x) satisfy the hypotheses of the MeanValue Theorem?

[0, 2] [0, 4] [3, 5] [4, 5] none of these

e. [2 points] On which of the following intervals does g(x) satisfy the conclusion of the MeanValue Theorem?

[0, 2] [0, 4] [3, 5] [4, 5] none of these

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 2 Problem 6 Solution


7. [14 points] The Kampyle of Eudoxus is a family of curves that was studied by the Greekmathematician and astronomer Eudoxus of Cnidus in relation to the classical problem ofdoubling the cube. This family of curves is given by

a2x4 = b4(x2 + y2).

where a and b are nonzero constants and (x, y) 6= (0, 0)—i.e.. the origin is not included.

a. [5 points] Finddy

dxfor the curve a2x4 = b4(x2 + y2).

Solution: Using implicit differentiation, we have

4a2x3 = 2b4x + 2b4ydy

dxso

dy

dx=

4a2x3 − 2b4x

2b4y.

b. [5 points] Find the coordinates of all points on the curve a2x4 = b4(x2 + y2) at which thetangent line is vertical, or show that there are no such points.

Solution: If the tangent is vertical, the slope is undefined. Setting the denominatorfrom part (a) equal to zero gives y = 0. Substituting y = 0 in the original equation gives

a2x4 = b4x2

and since (0, 0) is excluded, we know x 6= 0 so x2 =b4

a2, and x = ±b2

a.

Thus, there are two points on the curve where the tangent is vertical:

(

b2

a, 0

)

,

(

−b2

a, 0

)

.

c. [4 points] Show that when a = 1 and b = 2 there are no points on the curve at which thetangent line is horizontal.

Solution: Using a = 1 and b = 2 indy

dxfrom part (a), we have

dy

dx=

4x3 − 32x

32y.

If the tangent is horizontal, the slope is zero, so solving 4x3 − 32x = 4x(x2 − 8) = 0 givesx = 0 or x = ±

√8. We must see if any of these values of x give a point on the curve.

Note that when x = 0, y = 0—and this point has been excluded from the family. Ifx = ±

√8, the equation of the curve gives us 64 = 16(8) + 16y2, which gives y2 = −4, so

there is no such point on the curve. Thus, there are no horizontal tangents to the curvex4 = 16(x2 + y2).

University of Michigan Department of Mathematics Fall, 2009 Math 115 Exam 2 Problem 7 (Kampyle of Eudoxus) Solution


11. [5 points] A curve C gives y as an implicit function of x. The curve C passes through the point(1, 2) and satisfies

dy

dx=

y2 − 2xy + 4y − 5

4(y − x).

a. [1 point] One of the values below is the slope of the curve C at the point (1, 2). Circlethat one value.

Solution: Plugging x = 1 and y = 2 into the given formula for dydx

yields 3/4.

Answer: The slope at (1, 2) is 1

4

1

3

1

2

5

8

2

3

3

4

4

5

b. [4 points] One of the following graphs is the graph of the curve C.Which of the graphs I-VI is it? To receive any credit on this question, you must circleyour answer next to the word “Answer” below.

1

2

3

−1

−2

−3

1 2 3−1−2−3

x

y

I

1

2

3

−1

−2

−3

1 2 3−1−2−3

x

y

II

1

2

3

−1

−2

−3

1 2 3−1−2−3

x

y

III

1

2

3

−1

−2

−3

1 2 3−1−2−3

x

y

IV

1

2

3

−1

−2

−3

1 2 3−1−2−3

x

y

V

1

2

3

−1

−2

−3

1 2 3−1−2−3

x

y

VI

Solution: We know that the desired curve passes through the point (1, 2) with slope3/4. This allows us to eliminate Graph V (which doesn’t pass through (1, 2)) and GraphsII and VI (which have negative slope at (1, 2)).To decide between Graphs I, III, and IV, we look at other points on the graphs.Graph I passes through the point (2,−1) with negative slope, but the above formula fordydx

says that it should have positive slope there, so Graph I is incorrect.Graph III passes through the point (1,−2) with negative slope, but the above formulafor dy

dxsays that it too should have positive slope there, so Graph III is incorrect.

The only remaining possibility is Graph IV.(Note that we could have also eliminated all but Graph IV by checking for vertical tangentlines at points (x, y) with y = x.)

Remember: To receive any credit on this question, you must circle your answer next to theword “Answer” below.

Answer: I II III IV V VI



4. [8 points] A ship’s captain is standing on the deck while sailing through stormy seas. Therough waters toss the ship about, causing it to rise and fall in a sinusoidal pattern. Supposethat t seconds into the storm, the height of the captain, in feet above sea level, is given by thefunction

h(t) = 15 cos (kt) + c

where k and c are nonzero constants.

a. [3 points] Find a formula for v(t), the vertical velocity of the captain, in feet per second,as a function of t. The constants k and c may appear in your answer.

Solution: The velocity is the derivative of the height function, so we compute

v(t) = h′(t) = −15k sin (kt) .

Notice that the Chain Rule gives us a factor of k out front, and since c is an additiveconstant, it disappears when we take the derivative.Notice also that v(t) = dh

dtdoes indeed have units of feet per second, as required.

Answer: v(t) = −15k sin (kt)

b. [2 points] Find a formula for v′(t). The constants k and c may appear in your answer.

Answer: v′(t) = −15k2 cos (kt)

c. [3 points] What is the maximum vertical acceleration experienced by the captain? Theconstants k and c may appear in your answer. You do not need to justify your answer orshow work. Remember to include units.

Solution: The acceleration is just the derivative of the velocity function, which was justcomputed in the previous part.Since v

′(t) = −15k2 cos (kt) is sinusoidal with midline 0 and amplitude 15k2, the maxi-mum value it achieves is 15k2.Since v

′(t) = dv

dt, the units on the acceleration are feet per second per second, or feet per

second squared.

Answer: Max vertical acceleration: 15k2 ft/s2

University of Michigan Department of Mathematics Winter, 2014 Math 115 Exam 2 Problem 4 (ship) Solution

Math 115 / Exam 2 (Nov 11, 2014) page 10

9. [10 points] Our friend Oren, the Math 115 student, wants to minimize how long it will takehim to complete his upcoming web homework assignment. Before starting the assignment, hebuys a cup of tea containing 55 milligrams of caffeine.Let H(x) be the number of minutes it will take Oren to complete tonight’s assignment if heconsumes x milligrams of caffeine. For 10 ≤ x ≤ 55

H(x) =1

120x2−

4

3x+ 20 ln(x) .

Instead of immediately starting the assignment, he solves a calculus problem to determine howmuch caffeine he should consume.

a. [8 points] Find all the values of x at which H(x) attains global extrema on the interval10 ≤ x ≤ 55. Use calculus to find your answers, and be sure to show enough evidencethat the points you find are indeed global extrema.

Solution: Since H(x) is continuous on the interval 10 ≤ x ≤ 55, by the Extreme ValueTheorem, H(x) attains both a global minimum and a global maximum on this interval.These will occur at either endpoints or critical points.Now,

H′(x) =

x

60−

4

3+

20

x=

x2 − 80x+ 1200

60x=

(x− 60)(x− 20)

60x.

Thus, H(x) has exactly one critical point on the interval 10 ≤ x ≤ 55, and it is at x = 20.To determine the global extrema, we compare the values of H(x) at all critical pointsand endpoints

x 10 20 55

H(x) ≈ 33.55 ≈ 36.58 ≈ 32.02Thus, the global minimum is at x = 55, and the global maximum is at x = 20.

(For each answer blank below, write none in the answer blank if appropriate.)

Answer: global min(s) at x = 55

Answer: global max(es) at x = 20

b. [2 points] Assuming Oren consumes at least 10 milligrams and at most 55 milligrams ofcaffeine, what is the shortest amount of time it could take for him to finish his assignment?Remember to include units.

Solution: The minimum of H(x) occurs at x = 55, where H(55) ≈ 32.02.

Answer: ≈ 32 minutes

University of Michigan Department of Mathematics Fall, 2014 Math 115 Exam 2 Problem 9 (caffeine) Solution

Math 115 / Exam 2 (Nov 11, 2014) page 5

4. [12 points] Researchers are constructing a rectangular garden adjacent to their building. Thegarden will be bounded by the building on one side and by a fence on the other three sides.(See diagram below.) The fencing will cost them $5 per linear foot. In addition, they will alsoneed topsoil to cover the entire area of the garden. The topsoil will cost $4 per square foot ofthe garden’s area.Assume the building is wider than any garden the researchers could afford to build.

building

h feet (fencing)

w feet (fencing)

h feet (fencing) garden

a. [5 points] Suppose the garden is w feet wide and extends h feet from the building, as shownin the diagram above. Assume it costs the researchers a total of $250 for the fencing andtopsoil to construct this garden. Find a formula for w in terms of h.

Solution: A garden of these dimensions will require 2h+w feet of fencing and hw squarefeet of ground covered by topsoil. Thus,

250 = 5(2h+ w) + 4hw .

Solving for w we find

w =250− 10h

4h+ 5.

Answer: w =

250− 10h

4h+ 5

b. [3 points] Let A(h) be the total area (in square feet) of the garden if it costs $250 andextends h feet from the building, as shown above. Find a formula for the function A(h).The variable w should not appear in your answer.

(Note that A(h) is the function one would use to find the value of h maximizing the area.You should not do the optimization in this case.)

Solution: The area of the garden in square feet is given by hw. In part (a), a formulafor w in terms of h was found when h and w are the dimensions of a garden that will

cost $250 in supplies to construct. Thus, A(h) = h

(

250− 10h

4h+ 5

)

.

Answer: A(h) =h

(

250− 10h

4h+ 5

)

c. [4 points] In the context of this problem, what is the domain of A(h)?

Answer: 0 < h < 25

University of Michigan Department of Mathematics Fall, 2014 Math 115 Exam 2 Problem 4 (garden) Solution

5

4. (12 points) The zoo has decided to make the new octopus tank spectacular. It will be cylindricalwith a round base and top. The sides will be made of Plexiglas which costs $65.00 per squaremeter, and the materials for the top and bottom of the tank cost $50.00 per square meter. If thetank must hold 45 cubic meters of water, what dimensions will minimize the cost, and what is theminimum cost?

r

h

Let V denote the volume of the tank. Then V = πr2h = 45 m3.

Solving for h gives h =45

πr2m.

The area of the “sides” of the tank is 2πrh =90

rm

2, and the area of the circles for the top and

bottom of the tank is 2πr2.

Thus, the cost of the tank as a function of r is

C(r) = 65(90

r) + 50(2)(πr

2) =5850

r+ 100πr

2.

To minimize the cost, set C ′(r) = −

5850

r2+200πr equal to zero, since C ′(r) is defined for all values

of r in the domain (r > 0).

Solving for r, we get r3 =5850

200π, so r ≈ 2.104 meters.

Testing to see if this r value is in fact, the minimum, we use the second derivative test.

Since C ′′(r) = 200π + 25850

r3> 0 for all values of r in the domain, we see that C is concave up for

all values of r. Since C(r) → ∞ as r → 0 and as r → ∞, r = 2.104 is the global minimum of C.

radius ≈ 2.104 m

height ≈ 3.237 m

cost ≈ $4171

University of Michigan Department of Mathematics Winter, 2007 Math 115 Exam 3 Problem 4 (octopus2) Solution

University of Michigan Department of Mathematics Fall, 2002 Math 115 Exam 3 Problem 10 (trough) Solution


6. [11 points] On the axes provided below, sketch the graph of a single function y = g(x) satisfyingall of the following:

• g(x) is defined for all x in the interval −6 < x < 6.

• g(x) has at least 5 critical points in the interval −6 < x < 6.

• The global maximum value of g(x) on the interval −5 ≤ x ≤ −3 is 4, and this occurs atx = −4.

• g(x) is not continuous at x = −2.

• g′(x) (the derivative of g) has a local maximum at x = 0.

• g(x) is continuous but not differentiable at x = 1.

• g′′(x) ≥ 0 for all x in the interval 2 < x < 4.

• g(x) has at least one local minimum on the interval 4 < x < 6 but does not have a globalminimum on the interval 4 < x < 6.

• g(x) has an inflection point at x = 5.

Make sure your sketch is large and unambiguous.

Graph of y = g(x)

There are many possible solutions. One is shown below.

−5

−4

−3

−2

−1

1

2

3

4

5

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

x

y



6. [9 points] In each of the following problems, draw a graph of a function with all of the indicatedproperties. If there is no such function, then write “NO SUCH FUNCTION EXISTS”. Youdo not need to write any explanations. No partial credit will be given on each part of thisproblem.

a. [3 points] A continuous function f(x), whose domain is all real numbers, with the follow-ing four properties:(i.) f(x) attains a local minimum somewhere.(ii.) f(x) attains a local maximum somewhere.(iii.) f(x) does not attain a global minimum.(iv.) f(x) does not attain a global maximum.

Solution:

x

y

b. [3 points] A continuous function g(x), whose domain is the closed interval [0, 1], with thefollowing two properties:(i.) g(x) does not attain a global maximum on the interval [0, 1](ii.) g(x) attains a global minimum on the interval [0, 1].

Solution: NO SUCH FUNCTION EXISTS

c. [3 points] A differentiable function j(x) with the following two properties:(i.) The linear approximation to j(x) at x = 3 gives an overestimate when used to ap-proximate j(2).(ii.) The linear approximation to j(x) at x = 3 gives an underestimate when used toapproximate j(4).

Solution:

2 3 4x

y


math 115 — practice for exam 2 - university of …mconger/dhsp/115packet2soln.pdfhence x= 0.8,5...

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