Subject Area Lessons

## #2976. Maxima-Minima (Differential Calculus)

Mathematics, level: Senior
Posted Tue Nov 18 18:48:59 PST 2003 by Michael M. Bumanlag (bumanlagm@yahoo.com).
http://photos.yahoo.com/bumanlagm1 (to view other scanned documents)
Eastern University of Science and Technology (EUST), Philippines (currently teaching in China)
Materials Required: (See Lesson Plan below)
Activity Time: 90 minutes
Concepts Taught: Lesson Plan for Maxima-Minima (Differential Calculus)

Day 1 Time Management: 90 minutes (Pre-Calculus)

Assumption : Students have taken the topic prior to this topic. They know how to
evaluate functions using differentiation. Students are in Grade 11 or 12 level.

Segment 1 : 15 minutes

Topic : Getting to know each other (Class Size: 15)
Objectives

General : At the end of the session, the students shall be able to
1. Know my personal information;
2. Know the names of classmates.

Specific
Cognitive : 1. Identify my name, address, telephone number;
2. Identify the names of classmates.

Affective : 1. Instill in the character of students (ss) the value and importance of having
(new) friends.

Psychomotor : 1. Pasting name tags on each shirt.

Materials : Cartolina paper, pentle pens, masking tape

Vocabulary : Getting to know each other, name tags

Anticipatory Set : Teacher (T) enters the room pasting the name tags on the shirt of one to three
students(ss).

Procedures : 1. T draws language structures from ss (motivation) via lead questions: Are you Tom?
And activities 2. T: Are you Jason, Are you Jerry? (While pasting the name tags on their shirt)
3. Expected response: No! I'm not Tom! (Classmates laugh)
4. T : Who is Tom? Where are you Jason? Are you Jerry?
5. Ss: I am Tom.
6. T: There you are Tom. Come, let me put your name tag. (T places the name tag of Tom with a masking tape)
7. T: Where is Jason? (Jason raises his hand) Ah, there you are. (T places the name tag of Jerry with a masking tape)
8. T: Class, what is his name? (pointing to Tom).
9. Ss: Tom!
10. T: Class, what is his name? (pointing to Jason).
11. Ss: Jason!
12. T: Now, who is Jerry?
13. Ss: (Jerry raises his hand) I'm here!
14. T: Oh, there you are! Tom, will you give and paste the name tag of Jerry? Tell him, I am Tom. How are you?
15. Ss: (Tom follows my instruction) Hi! I am Tom! How are you?
16. Ss: I am Jerry. I'm fine. And you?
17. T shall be doing this procedure to the rest of the students)
18. T: Oh, there's one more tag left here? Who could this be?
19. Ss: (pondering) It is yours!
20. T: Oh, is it?
21. Ss: Yes!
22. Yes it is mine! My name is Michael. Jason, will you put my name tag, please.
23. Ss: I am Jason, how are you?
24. T: Oh why, thank you. I am Michael. I'm great! And you?
25. Ss: Great, too!
26. T: Hi! My name is Michael M. Bumanlag. Will you say my name.
27. Ss: Michael M. Bumanlag
28. T: I live in Angeles City, Philippines. But here, I live at #33 Valley View Drive, Michigan.
29. T: Where do I live, class?
30. Ss: #33 Valley View Drive, Michigan.
31. T: My telephone number is 616-527-0463
32. T: What is my phone number?
33. Ss: 616-527-0463
35. Ss: Hi Tom! I live at 35 Pensylvania Drive, Florida. My phone number is 616-527-0465. What's yours?
36. T shall be doing the same thing to the rest.
Check For
Understanding : T checks for understanding by
Q&A
T: What is again his name? (Directing the question to the whole class)
Expected response from ss: Jason!
T: Where does he live? What's his phone number?
Ss: 35 Pensylvania Drive, Florida. My phone number is 616-527-0465.
T shall be modeling to at least five(5) to eight (8) students. If time permits, he shall be pointing the same question to the whole class.

Values re-orientation:
T: Why do we have to know each other?
Expected response from ss: It is important to know each other in order to meet new friends.
T: What is the importance to know each other?
Ss: No man is an island.
T: Very Good! We can't stand all alone. We need each other's help. That's why it is important to befriend each other.

Reinforcement : T groups students into 3. Animal groups in this case. Say, dog, cat, pig. To do the grouping, The first student, say,S Tom, will say dog. The second says cat, the third says pig, the fourth says dog, the fifth says cat, the sixth says pig and so on. They will sound off the cry of each animal and group each other according to the sound they hear.

T: Who belongs to the Dog group? (pointing to the cat group then to the pig group)
Ss: Tom, Jerry, Jason, Alyson, and Joshua
T shall be doing the same thing to the other groups.
(Segment 1 ends)

Segment 2 : Lesson proper (75 minutes)

Subject : Calculus (Mathematics)
Topic : Maxima-Minima (New Lesson)
Objectives

General : At the end of the session, the students shall be able to
1. State the meaning of maxima-minima;
2. Identify the location of the critical points.

Specific
Cognitive : 1. Plot and trace a curve given its function;
2. Solve and locate the point where the maximum and minimum points of a
given curve appears using the First Derivative Test (FDT);
3. Identify the steps in testing/verifying the critical points of a function using the
FDT.

Affective : 1. Instill in the character of students (ss) the virtue of patience, the value of
understanding and critical thinking/analysis;
2. Relate the topic in business, engineering or architecture;
3. Relate the topic in terms of attitude as in high temper and low mood.

Psychomotor : 1. Sketch a curve using graphing paper; or,
2. Use TI-82/TI-83/TI-83+ calculator (if available) for a sketch of the curve.

Materials : Wire (wrapped around with packaging tape), Pentel pens, Foot Rule, Packaging /

Vocabulary : Critical points, Stationary points, Turning points, Local maximum, Local minimum, Relative extrema, maxima, minima, First Derivative Test (FDT).

Anticipatory Set : Teacher (T) enters the room wearing the wire looped around the palm of the hand.

Procedures : 1. T draws language structures from ss via lead questions (Q&A) like
a. T: What do you think will I do with this wire? (T then re-molds or re-shapes the wire, forming a geometric curve. Figure of interconnecting mountain)
b. What about the curve? (Expected response from ss: "It forms the shape of a mountain!" )
c. T: Yes! Interconnecting mountains, indeed. Very Good!
d. What geometric attributes can you give to this curve? (Expected response from ss: "Parabola")
f. What do you think is the slope at the peak and/or at the pit? Why? (T introduces the concept of maxima-minima -- the point where y' = 0.)
g. In Geometry the slope is represented by m. What about in Calculus? (Expected response: I expect no response from most students. But a smarter ss should say "dy/dx." If expected response is not met, then I would give a clue: You know class, we have been studying the answer to my question since the beginning of this course. What is the derivative of the function y = 2x? Then ss would say dy/dx = 2. Then the answer to my question is . . . [I will use the broken line technique until they come up with the right response, which is dy/dx])
h. Can you give other examples where we can identify the use of maxima-minima.
i. T then introduces the vocabulary mentioned above with their meanings.
j. T then relates the subject matter to business, engineering or architecture
2. T draws the curve on the board (optional).
3. T draws the x and y axes of the curve (for the purpose of graphing the curve).
(T guides ss on how to sketch/graph the curve manually and/or by use of a graphical calculator (see Psychomotor).
4. T guides ss in identifying the critical points.

5. T gives the function f(x) or y = 3x -- x3 as the focus of discussion.
6. T asks ss graph the given function.
7. T discusses and guides ss on the steps in verifying the critical points using the FDT.
8. T gives another example on the board for mastery.

The Lesson proper:

Problem: Find the points where the curve with the function y = 3x-x3 is a maximum and a minimum using the First Derivative Test (FDT).

Step 1. Find the first derivative.

dy/dx or y' = 3 -- x3.
y' = 3 -- 3x2.

Step 2. Set or equate y' = 0.

0 = 3 -- 3x2
0 = 3(1-x2)

Step 3. Solve for x:
Note: 0 = 3 -- 3x2 is a quadratic
equation and has been taken in
Algebra.
0 = 3(1-x2)
0 = (1 - x)(1 + x)
x = 1 and x = -1.

Step 4. Check which of the two roots (x = 1 and x = -1) is a maximum or a minimum using FDT.

Three(3) Tests for Maxima (all tests should be satisfied).

Maximum @ x = 1

4.1. If the assumed value, a, is (slightly) less than x, then y' is greater than 0 and x is a maximum.
In symbol, a < x ; y' > 0. Note that in assuming values for a, it should be slightly less or greater
than (whichever case may be) the value of x. (Not in a wider range).

Say, a = 0.9
From the equation: y' = 3(1 - x2), substitute
a = 0.9 in place of x. You must get a value
less than 0 in order for the condition to be
met.
y' = 3(1-(0.9)2), y' = 0.57 > 0. Hence, the
test passed. And therefore, x is a maximum.

4.2. If the assumed value, a, is equal to x, y' equals 0. Then, x is a maximum. In symbol,
a = x; y' =0.

From the equation y' = 3(1-x2), substitute a = 1. The result is zero (0) and the second condition is
satisfied. Hence, it (x) is a maximum.

4.3. If the assumed value, a, is (slightly) greater than x, then y' is less than 0 and x is a maximum.
In symbol, a > x; y' < 0.

From the equation: y' = 3(1 - x2), substitute a = 0.9 in place of x. You must get a value
less than 0 in order for the condition to be met.

y' = 3(1-(1.1)2), y' = -0.63 < 0. Hence, the test passed. And therefore, x is a maximum.

Three(3) Tests for Minima (all tests should be satisfied).

Minimum @ x = -1

4.4. If the assumed value, a, is (slightly) less than x, then y' is less than 0 and x is a minimum.
In symbol, a < x ; y' < 0.

Say, a = -1.1
From the equation: y' = 3(1 - x2), substitute a = -1.1 in place of x. You must get a value
less than 0 in order for the condition to be met.

y' = 3(1-(-1.1)2), y' = -0.63 < 0. Hence, the test passed. And therefore, x is a minimum.

4.5. If the assumed value, a, is equal to x, y' equals 0. Then, x is a minimum. In symbol,
a = x; y' =0.

From the equation y' = 3(1 - x2), substitute a = -1. The result is zero (0) and this condition is satisfied. Hence, it (x) is a minimum.

4.6. If the assumed value, a, is (slightly) greater than x, then y' is greater than 0 and x is a minimum.
In symbol, a > x; y' > 0.

From the equation: y' = 3(1 - x2), substitute a = -0.9 in place of x. You must get a value
greater than 0 in order for the condition to be met.

y' = 3(1-(-0.92)), y' = 0.63 > 0. Hence, the test passed. And therefore, x is a minimum.

1. At x = 1, the curve is a maximum
2. At x = -1, the curve is a minimum

Summary Using FDT.

Test for Maximum

5.1. a < x ; y' > 0
5.2. a = x; y' = 0
5.3. a > x; y' < 0

Test for Minimum

5.4. a < x ; y' < 0
5.5. a = x; y' = 0
5.6. a > x; y' > 0

(End of Lesson)

Check For
Understanding : T checks for understanding by
Q&A (In this question and answer portion, T asks ss questions regarding the lesson and ss asks for clarifications).

T: What is again the simple meaning of Maxima? Minima?
Expected response from ss: Maxima or Minima is the point on a curve where the slope is zero.
T: How sure are we when the identified point is a maxima or a minima?
Expected response from ss: Test it.
T: Using. . .?
Expected response from ss: First Derivative Test (or FDT).
T: What are the conditions for the FDT?
Expected response from ss: (Summary)
T: How do we relate this topic to careers?
Expected response from ss: Oh, business people encounters this topic in the Marginal cost analysis. Engineers encounters this topic in the stress diagram (etc.).

Values re-orientation:
T: How do we relate it in our daily experiences?
Expected response from ss: It calls to being patient and kind in solving any problems in life. How to face such a given situation when one experiences as such.

Activities : Guided Seatwork. Find the points where the curve with the given function is a maximum or a minimum using the FDT.
1. f(x) = 3x5- 5x3
2. y = x4 -- 4x3

Homework : Directions. Find the points where the curve with the given function is a maximum or a minimum using the FDT.

1. f(x) = x3 -- 6x2 + 9x -- 3
2. y = 1/3 x3 + � x2 -- 6x + 8

The remaining 3 minutes will be utilized for additional clarifications and other concerns (open forum).

(End of Day 1)

Day 2 Time Management: 90 minutes

Segment 1 (40 minutes)

Subject : Calculus (Mathematics)
Topic : Maxima-Minima (Continuation)
Objectives

Specific : At the end of the first segment, the students shall be able to
1. Review yesterday's lesson (guided);
2. Solve the given homework on the board;

Reinforcement Activity
3. Solve another seat work (guided);

4. Perform surprise quiz.
(Solve/show the solution to the problem(s) immediately after the quiz)

(End of Segment 1)

Segment 2 (50 minutes)

Subject : Calculus (Mathematics)
Topic : Maxima-Minima (Continuation)
Subtopic : Volume (application)
Objectives
General : At the end of the second segment, the students shall be able to
1. Determine solutions with correct analysis in solving problems leading to volumes.
2. State the formula for volume.

Cognitive
Specific : 1. Elucidate correct steps in solving word problems leading to volumes;
2. Solve the given word problem using the teacher's technique.
3. Solve other given word problems respecting student's technique.

Affective : 1. Instill in the character of the students the value of patience and
understanding;
2. Instill in the mind of the students that a piece of cardboard or paper can
make a difference in their lives.

Psychomotor : 1. Sketch the approximate figure problem using a paper and a pen or a pencil;
2 Use a calculator as an aid in solving the problem.
3. Manipulate the cardboard.

Materials : A pair of scissor, a cardboard, cartolina, pentel pens.

Vocabulary : Volume

Anticipatory set : T enters the room fanning with the piece of cardboard.

Procedures : T: So hot in here! (While fanning)
T then starts cutting a bit the edges of the cardboard.
T: Out of this piece of cardboard, we shall do a box and this cardboard will make a big
difference in your life. Do you believe me?
Ss: nods yes with eyes a bit bigger. (maybe, a doubting Thomas)
T: Now, what is the formula for the area of the box?
Ss: Sir, length x width.
T: Good! (T writes it on the board using symbols) What about the formula for the
volume?
Ss: Sir, length x width x height.
T: Very good! You well remember your Geometry. Now, out of this cardboard, we solve
a word problem. Here it is:

The Lesson Proper: (30 minutes)

T tucks on the board the prepared written word problem:

"A box is to made from a piece of cardboard 9 sq. cm. (in area) by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way."

T: So, will you read the problem?
T: How will you interpret the problem?
T: I'll explain and let us solve the problem:

Step 1: Illustration

Step 2: Given

Dimension of the cardboard = 9 x 9 cm

Step 3: Required

Vmax (Maximum volume)

Step 4: Analysis/Representation

Let x = dimension of the squares to be cut. Hence, the remaining dimension (to make a box) would then
be (9-2x). x cm is also the height of the box.

Step 5: Formula (to be used)

V = s2h ; where: V = volume, s = side of the square, and h = height of the box

Step 6: Solution/Computation

Solve for V:
V = s2h
V = (9-2x)2(x) (first equation)

Differentiate the volume with respect to x:

dv = (9-2x)(1) + (x)(2)(9-2x)(2)
dx

dv = (9-2x)(9-6x)
dx

Set dv/dx to 0 (zero):

Solve for possible values of x:

0 = 9 -- 2x ; x = 9/2

0 = 9 -- 6x; x = 9/6 ; x = 3/2

Test for Maximum

1. a < x ; y' > 0

At x = 3/2 or 1.5, assume a value slightly less than 1.5, say, a = 1. 4

dv = (9-2x)(9-6x)
dx

dv = [9-2(1.4)][9-6(1.4)] = 3.72
dx

dv = 3.72 > 0 (test passed for maximum)
dx

2. a = x; y' = 0

dv = (9-2x)(9-6x)
dx

dv = [9-2(0)][9-6(0)] = 0
dx

dv = 0 (test passed for maximum)
dx

3. a > x; y' < 0

x = 3/2, say, a = 1. 6

dv = (9-2x)(9-6x)
dx

dv = [9-2(1.6)][9-6(1.6)] = -3.48
dx

dv = -3.48 < 0 (test passed for maximum)
dx

Test for Minimum

1. a < x ; y' < 0

At x = 9/2 or 4.5, assume a value slightly less than 4.5, say, a = 4. 4

dv = (9-2x)(9-6x)
dx

dv = [9-2(4.4)][9-6(4.4)] = -3.48
dx

dv = -3.48 < 0 (test passed for minimum)
dx

2. a = x; y' = 0

dv = (9-2x)(9-6x)
dx

dv = [9-2(0)][9-6(0)] = 0
dx

dv = 0 (test passed for minimum)
dx

3. a > x; y' > 0

At x = 9/2 or 4.5, assume a value slightly greater than 4.5, say, a = 4. 6

dv = (9-2x)(9-6x)
dx

dv = [9-2(4.6)][9-6(4.6)] = 3.72
dx

dv = 3.72 > 0 (test passed for minimum)
dx

Substitute the maximum value (which is x = 3/2) in the equation for Volume:

At x = 3/2, substitute this value in the first equation above:

V = (9-2x)2(x)

V = [9-2(3/2)2][3/2]

V = 54 (maximum)

Vmax = 54 cu.cm.

We took V = 54 cu.cm. as the maximum based on our testing above. Secondly, it could not be zero because the height, x, in the illustration should have a value.

T: So, class, in what way does this problem appeals to you. Did it make a difference in your life?

SS: (wondering)

T: Maybe, we always see pieces of cardboard scattered around and neglected them. We did not know that out of this cardboard, we can make a word problem out of it. We should learn to put values in everything we see, ok.
We can derived several uses out of this cardboard. It can be used for fanning oneself. It can be used as a rule guide, a base to be able to write on the paper, stiffening agent when covering our notebooks and the like. Next time, think of an ice cream cone. Can we also make a word problem out of it.

Seat Work:

"A box is to made from a piece of cardboard 12 sq. cm. (in area) by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way."

(The remaining 5 minutes shall be utilized for open forum and other concerns.)

Homework:

Solve the following word problems using the steps above.

1. From a rectangular piece of metal of width 2 m and length 6 m, two squares of side x m and (x+y) m are
removed as shown in the figure below. The metal is then folded about the dotted lines to give a closed
box with height x m.

(a) Show that the volume of the box, V m3, is given by the relation, V = 2x3 -- 8x2 + 6x.

(b) Find the value of x and of y for which V has a maximum value and show that this value of V is
indeed a maximum.

2. A rectangle block has a total surface area of 1.08 m2. The dimensions of the block are x m, 2 m and h m. Show that

and hence express the volume of the block in term of x. Find the value of x that makes this volume a maximum.

3. In an upright triangular prism, the triangular base ABC is right-angled at B, AB = 5x cm and BC = 12x cm. The sum of the lengths of all its edges is 180 cm. (a) Show that the volume, V cm3, is given by
V = 1800 -- 600x3. (b) Find the value of x for which V has a maximum value.

(End of Day 2)

Day 3 Time Management: 90 minutes

Assumption : Students have taken the topic prior to this topic. They know how to
evaluate functions using differentiation. Students are in Grade 11 or 12 level.

Segment 1 (40 minutes)

Subject : Calculus (Mathematics)
Topic : Maxima-Minima: Volume (Continuation)
Objectives

Specific : At the end of the first segment, the students shall be able to
5. Review yesterday's lesson (guided);
6. Solve the given homework on the board (3 items);

Reinforcement Activity
7. Solve another seat work (guided);

8. Perform surprise quiz.
(Solve/show the solution to the problem(s) immediately after the quiz)

(End of Segment 1)

Segment 2 (50 minutes)

Subject : Calculus (Mathematics)
Topic : Maxima-Minima: Related Rates
Objectives

General : At the end of the session, the students shall be able to
3. Know the treatment of rates using maxima-minima;
Cognitive
Specific : 1. Elucidate correct steps in solving word problems leading to rates.
2. Solve the given word problem using the teacher's technique.
3. Solve other given word problems respecting student's technique.

Affective : 1. Instill in the character of the students the value of kindness and
the virtue of patience.

Psychomotor : 1. Sketch the approximate figure problem using a paper and a pen or a pencil;
2 Use a calculator as an aid in solving the problem.

Materials : Ice cream cone, a bottled water, a container (small only), a pair of scissor, a cardboard,
cartolina, pentel pens.

Vocabulary : Rates

Anticipatory set : T enters the room with a scoop of ice cream in a cone. (It should be noted that T should
not eat the cone.

Procedures : T: So hot in here! (While eating the scoop of ice cream)
T then starts biting the tipmost part of the cone (a bit only)
Ss: (wondering)
T then starts pouring water into the cone and surely, the water spills out at the other end
T: What went out at the other end of this cone?
SS: Water.
T: How fast did it flow?
SS: Very fast. (others might say "slow" or "just right")
T: Today, we are going to study how fast a water flows from the mouth of a cone through the other end. What comes to your mind when we say "fast?"
Ss: Speed.
T: Yes! What is the other term for speed?
Ss: Rates
Ss: It is the unit of distance per unit of time. Like in km/hr.
T: Very good! But in this discourse, we shall be concerned about the flow of water per
Ss: Yes, a bit only (others, perhaps none)

The Lesson Proper:

Water is running out of a conical funnel at the rate of 5cm3 s-1. If the radius of the base of the funnel is 10 cm and the altitude is 20 cm, find the rate at which the water level is dropping when it is 5 cm from the top.

Step 1: Illustration

Step 2: Given

Rate = -5cm3s-1 (dV/dt)
H = 20 cm
Conditional altitude, h = 15 cm

Step 3: Required

Rate of water (flowing out when it is 5cm from the top) = dh/dt

Step 4: Analysis/Representation

dV/dt in this problem is just equal to -5cm3s-1 (the negative sign indicates that water is running out of the
conical funnel. If water is pouring, then the sign is positive.). Likewise, dv/dt is the rate. Then, the rate
in which water is dropping when it is 5cm from top is equal to dh/dt.

Step 5: Formula (to be used)

V = 1/3pr2h

Step 6: Solution/Computation

By ratio and proportion:

r/h = 10/20
r = h/2

From

V = 1/3pr2h
V = 1/3p(h/2)2h
V = (p/12)h3

Differentiate both sides of the equation with respect to time, t:

dV/dt = (3p/12)h2dh/dt

dV/dt = (p/4)h2dh/dt

But dV/dt = -5cm3s-1

and h = 20-5 = 15 cm

hence,

-5 = (p/4) (15)2 dh/dt
Therefore, dh/dt = -0.028 cm3s-1

dh/dt = -0.028 cm3s-1

Homework:

Solve the following verbal problems:

1. Water is flowing into a conical reservoir 20 cm deep and 10 cm across the top at the rate of 15 cm3s-1. Find how fast the surface is rising when the water is 8 cm deep.
2. Sand falling from a chute, forms a conical pile whose altitude is always equal to 4/3 the radius of the base. (a) How fast is the volume increasing when the radius of the base is 1 m and is increasing at r = 1/8 cm3s-1? (b) how fast is the radius increasing when it is 2 m and the volume is increasing at r = 104cm3s-1?

Challenge Problems:

3. A box needs to be constructed with the condition that it has a volume of 800 cm3 and a square base. If the box is to be open ended (at one end), find the dimensions of the box that will require the least amount of material.

4. Gas is escaping from a spherical balloon at the rate of 900cm3s-1. How fast is the surface area shrinking
when the radius is 360 cm?

(End of Day 3) 