Basic Calculus: Quarter 3 - Module 9: Optimization Problems
Basic Calculus: Quarter 3 - Module 9: Optimization Problems
Basic Calculus: Quarter 3 - Module 9: Optimization Problems
Quarter 3 – Module 9:
Optimization Problems
Basic Calculus – Grade 11
Alternative Delivery Mode
Quarter 3 – Module 9: Optimization Problems
First Edition, 2020
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Team Leaders:
School Head : Marijoy B. Mendoza, EdD
LRMDS Coordinator : Karl Angelo R. Tabernero
Each SLM is composed of different parts. Each part shall guide you step-by-
step as you discover and understand the lesson prepared for you.
In addition to the material in the main text, Notes to the Teacher are also
provided to our facilitators and parents for strategies and reminders on how they can
best help you on your home-based learning.
Please use this module with care. Do not put unnecessary marks on any part
of this SLM. Use a separate sheet of paper in answering the exercises and tests. And
read the instructions carefully before performing each task.
If you have any questions in using this SLM or any difficulty in answering the
tasks in this module, do not hesitate to consult your teacher or facilitator.
Thank you.
What I Need to Know
One of the main reasons why this module was created is to ensure that it will assist
you to understand the concept and know how to solve optimization problems.
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What I Know
1. Look for two positive numbers whose product is 36 and whose sum is a minimum.
A. 9 𝑎𝑛𝑑 4 B. 6, 𝑎𝑛𝑑 6 C. 18 𝑎𝑛𝑑 2 D. 8 𝑎𝑛𝑑 8
2. The sum of two positive numbers is 50. Find these two numbers whose product is
a maximum.
A. 20 𝑎𝑛𝑑 30 B. 40 𝑎𝑛𝑑 10 C. 25 𝑎𝑛𝑑 25 D. 24 𝑎𝑛𝑑 26
3. Look for two positive numbers whose product is 400 and whose sum is a
minimum.
A. 100 𝑎𝑛𝑑 4 B. 50 𝑎𝑛𝑑 8 C. 20 𝑎𝑛𝑑 20 D. 200 𝑎𝑛𝑑 2
4. The sum of two positive numbers is 60. Find these two numbers whose product is
a maximum.
A. 50 𝑎𝑛𝑑 10 B. 20 𝑎𝑛𝑑 40 C. 35 𝑎𝑛𝑑 25 D. 30 𝑎𝑛𝑑 30
5. Look for two positive numbers whose sum is 80 and whose product is a maximum.
A. 40 𝑎𝑛𝑑 40 B. 20 𝑎𝑛𝑑 60 C. 45 𝑎𝑛𝑑 35 D. 30 𝑎𝑛𝑑 50
6. Solve for the dimensions of the rectangle whose perimeter is 120 m and yields a
maximum area.
A. 50𝑚 𝑎𝑛𝑑 10𝑚 B. 20𝑚 𝑎𝑛𝑑 40𝑚 C. 35𝑚 𝑎𝑛𝑑 25𝑚 D. 30𝑚 𝑎𝑛𝑑 30𝑚
8. Rex inherits a piece of land. He wants to install fences on its rectangular region
and maximize the area. He already bought 320 m of fencing materials. Find the
optimized dimensions of the rectangular region.
A. 50𝑚, 10𝑚 B. 20𝑚, 40𝑚 C. 80𝑚, 80𝑚 D. 30𝑚, 30𝑚
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9. A house owner plans to build a rectangular floral garden beside her wall. Since
the wall already exists, only 3 sides of the rectangle have to be provided with
fence. The fencing materials were already purchased and have a quantity of
240m. Solve for the dimensions of the fence with the largest area possible.
A. 100𝑚 𝑎𝑛𝑑 30𝑚 B. 20𝑚 𝑎𝑛𝑑 90𝑚 C. 120𝑚 𝑎𝑛𝑑 60𝑚 D. 30𝑚 𝑎𝑛𝑑 30𝑚
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Lesson
1 Optimization Problems
Nowadays, you cannot be a big spender or too thrifty because somehow it will have
an unwanted impact on a fixed monthly budget. Most people prefer to be more in
control of the things they want to purchase because they want to maximize the value
of their money. This lesson can help people optimize certain situations which lead to
a better decision.
What’s In
What’s New
Solve the given problem. Use the fundamental step procedure in solving word
problems.
Recall: What is asked? What are the given? What fundamental operation to be used?
Give the solution and final answer.
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What is It
• Optimization problems are word problems that deal with the application of
finding the maximum or minimum value of a function. It has a constraint and
an equation that needs to be optimized.
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Example 1
Solve for two positive numbers whose product is 121 and whose sum is a
minimum.
Solution:
Let 𝑥 and 𝑦 be the two positive missing numbers with a product of 121.
𝑥 ∙ 𝑦 = 121 𝑀 =𝑥+𝑦
1. List what is asked on the
problem. Constraint Optimization
2. Write the constraint and the equation equation
optimization equation. (minimized)
121
3. On the constraint equation, 121 𝑀 =𝑥+( )
solve for 𝑦. Then, substitute that 𝑦= 𝑥
𝑥 −1
equation to its corresponding 𝑦 on 𝑀 = 𝑥 + 121𝑥
the optimization equation.
𝑀 ′ = 1 + (−1)(121𝑥 −2 )
4. Simplify and take its first
derivative. 121
𝑀′ = 1 − ;
𝑥2
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7. Substitute the minimum 𝑥 value If 𝑥 = 1, then
to the simplified constraint
equation to solve for 𝑦. 121
𝑦= = +11
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Example 2
Solution:
Let 𝑥 and 𝑦 be the missing sides of the rectangle with perimeter 𝑃 = 250𝑚.
2𝑥 + 2𝑦 = 250 𝐴=𝑥∙𝑦
2. Write the constraint and the
optimization equation.
Optimization
Constraint
equation
equation
(maximized)
3. On the constraint equation,
solve for 𝑦. Then, substitute that 𝑦 = 125 − 𝑥 𝐴 = 𝑥 ∙ (125 − 𝑥)
equation to its corresponding 𝑦
on the optimization equation.
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6. Test the 𝑥 value. Substitute Test:
it to the second derivative and
check whether the answer is less 𝐴′ = 125 − 2𝑥
than or greater than zero. Since
𝐴′′ = −2 𝐴′′ < 0
the answer is less than zero,
then it is the maximum value.
121 ′′
𝑀′′ = 𝑀
7. Substitute the maximum If 𝑥 = 𝑥 3 𝑚, then
62.5
121
𝑥 value to the simplified = 3
constraint equation to solve
𝑦 = 𝑥125 − (62.5) = 62.5𝑚
for 𝑦.
Example 3
An illustration board with 12 cm by 12 cm dimensions has to be made as a
square box with no top cover. Square corners should have equal cut outs to
enable folding of every side. Find the dimensions of the box that will give the
maximum volume.
Solution
Let 𝑥 be the height of the box, (12 − 2𝑥) be the length, and (12 − 2𝑥) be the width of
the box.
1. Draw a diagram. List what is asked on the problem and label the
diagram with relevant data.
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2. Write the constraint and the Constraint equations
optimization equation.
𝑙 = 12𝑥 − 2𝑥
𝑤 = 12𝑥 − 2𝑥
ℎ=𝑥
𝑠𝑖𝑑𝑒𝑠 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 [0, 6]
6𝑥 12 2𝑥 12
= =
6 6 2 2
𝑥=𝟐 𝑥=𝟔
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7. Test the 𝑥 value. Substitute Test:
it to the second derivative and
check whether the answer is less
𝑉 ′ = 12𝑥 2 − 96𝑥 + 144
than or greater than zero. Since 𝑉 ′′ = 24𝑥 − 96
the answer is less than zero,
n 𝑉 ′′ = 24(2) − 96 = −48
then it is the maximum value.
𝑉 ′′ < 0
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What’s More
Solve the following word problems by completing the table below. Fill-out the right
column with corresponding solutions by following the steps on the left column to
arrive at the final answer. Copy and answer the right column on a separate sheet of
paper.
1. Provide two positive numbers with a product of 169 and a sum which is a
minimum.
Steps Solution
Constraint equation:
1. List what is asked on the problem.
7. Substitute the minimum 𝑥 value to 𝑥 = _______ and 𝑦 = _______ are the two
the simplified constraint equation to positive numbers
solve for 𝑦.
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2. Determine the dimensions of a rectangle whose perimeter is 300 m with a
maximum area.
Steps Solution
Constraint equation:
12
3. An illustration board with 36 cm by 36 cm dimensions has to be made as a
square box with no top cover. Square corners should have equal cut outs to
enable folding of every side. Find the dimensions of the box that will give the
maximum volume.
Steps Solution
Constraint equation:
ℎ𝑒𝑖𝑔ℎ𝑡 = ________
9. Solve for the dimensions. 𝑙𝑒𝑛𝑔𝑡ℎ = ________
𝑤𝑖𝑑𝑡ℎ = ________
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What I Have Learned
Find out what you have learned in this lesson by answering the questions below on
a separate sheet of paper.
3. Mention past topics in mathematics that you applied on the last activity to solve
optimization problems.
What I Can Do
A resort owner wants to enclose a beachfront area for swimming activities. Based on
her plan, only 3 sides will be fenced with 270-meter rope and floats, while the
shoreline part will be open. Determine the dimensions of the 3 sides of the rectangle
that will give a maximum area.
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Assessment
Solve the following word problems. Write your answer on a separate sheet of paper.
1. Look for two numbers with a difference of 6 such that the product is a minimum.
A. 10 𝑎𝑛𝑑 4 B. 12, 𝑎𝑛𝑑 6 C. 3 𝑎𝑛𝑑 − 3 D. 2 𝑎𝑛𝑑 8
2. Variables 𝑎 and 𝑏 represent two positive numbers such that 𝑎 + 𝑏 2 = 1. Find these
two numbers that will maximize the equation, 𝑃 = 𝑎𝑏 2 .
1 √3 1 √2 1 √5 1 √7
A. 𝑎 = , 𝑏 = B. 𝑎 = , 𝑏 = C. 𝑎 = , 𝑏 = D. 𝑎 = , 𝑏 =
3 2 2 2 4 2 6 2
3. Look for two numbers with a difference of 12 such that the product is a minimum.
A. 6 𝑎𝑛𝑑 − 6 B. 12, 𝑎𝑛𝑑 24 C. 3 𝑎𝑛𝑑 15 D. 7 𝑎𝑛𝑑 8
4. Variables 𝑎 and 𝑏 represent two positive numbers such that 𝑎 + 𝑏 2 = 4. Find these
two numbers that will maximize the equation, 𝑃 = 𝑎𝑏 2 .
A. 𝑎 = 2, 𝑏 = √2 B. 𝑎 = 3, 𝑏 = √2 C. 𝑎 = 2, 𝑏 = √3 D. 𝑎 = 4, 𝑏 = √2
5. Find two numbers with a sum of 5 so that the equation 𝑆 = 2𝑥 2 + 3𝑦 2 will give a
minimum value.
A. 6 𝑎𝑛𝑑 − 6 B. 3 𝑎𝑛𝑑 2 C. 1 𝑎𝑛𝑑 4 D. −2 𝑎𝑛𝑑 8
6. Solve for the dimensions of the rectangle with a perimeter of 450 m such that the
area is a maximum.
A. 50𝑚 𝑎𝑛𝑑 10𝑚 C. 35𝑚 𝑎𝑛𝑑 25𝑚
B. 20𝑚 𝑎𝑛𝑑 40𝑚 D. 112.5𝑚 𝑎𝑛𝑑 112.5𝑚
7. Find two numbers with a sum of 3 so that the equation 𝑆 = 𝑥 2 + 2𝑦 2 will give a
minimum value.
A. 6 𝑎𝑛𝑑 − 5 B. 1, 𝑎𝑛𝑑 4 C. 3 𝑎𝑛𝑑 1 D. 2 𝑎𝑛𝑑 1
8. Lito wants to fence his subdivision lot. He wants to install fences on its
rectangular region and maximize the area. He measured and found out that its
perimeter is 280 m. What are the dimensions of the rectangular region that will
give him the maximum area possible?
A. 120𝑚 𝑎𝑛𝑑 10𝑚 B. 25𝑚 𝑎𝑛𝑑 80𝑚 C. 35𝑚 𝑎𝑛𝑑 25𝑚 D. 70𝑚 𝑎𝑛𝑑 70𝑚
15
9. A copper sheet with 16 cm by 16 cm dimension has to be made as a square box
with no top cover. Square corners should have cut outs to enable folding of every
side. Find the dimensions of the box that will give the maximum volume.
32
A. 𝑙 𝑎𝑛𝑑 𝑤 = 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 = 8/3𝑐𝑚
3
16 4
B. 𝑙 𝑎𝑛𝑑 𝑤 = 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 =
3 3𝑐𝑚
7
C. 𝑙 𝑎𝑛𝑑 𝑤 = 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 = 5/3𝑐𝑚
3
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D. 𝑙 𝑎𝑛𝑑 𝑤 = 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 = 2/3𝑐𝑚
3
Additional Activities
A rectangular box with open top and has a square base is a school project of two
senior high school students. However, they are wondering what would be the
minimum dimensions of their box if it will have a fixed volume of 62.5 cm³.
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What’s New
What is asked: Area of the rectangular basketball
court
What are the given: length: 20m width:10m
Formula: A = L x W
Solution: A = 20m x 10m = 200m
What’s In
1. A = 35m
2. P = 16m
3. x = 3/2, y = 5/2
4. y’ = 2x+7
What I Know
1. B
2. C
3. C
4. D
5. A
6. D
7. B
8. C
9. C
10. A
Answer Key
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Additional Activity
Length=width = 5m
Height= 2.5 m
What I Know
1. B
2. C
3. C
4. D
5. A
6. D
7. B
8. C
9. C
10. A
What I Can Do
x = 135m, y = 67.5m
What’s More
1. x = 13, y = 13
2. x = 37.5m, y = 37.5m
3. length=width = 24, height = 6
References
DepEd. 2013. Basic Calculus. Teachers Guide.
Lim, Yvette F., Nocon, Rizaldi C., Nocon, Ederlina G., and Ruivivar, Leonar A. 2016.
Math for Engagement Learning Grade 11 Basic Calculus. Sibs Publishing House,
Inc.
Mercado, Jesus P., and Orines, Fernando B. 2016. Next Century Mathematics 11
Basic Calculus. Phoenix Publishing House, Inc.
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