Basiccalculus q3 mod9 optimizationproblems final-STEM 12 SHS PDF

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Basic CalculusQuarter 3 – Module 9:Optimization ProblemsBasic Calculus – Grade 11 Alternative Delivery Mode Quarter 3 – Module 9: Optimization Problems First Edition, 2020Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However,...

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Basic Calculus Quarter 3 – Module 9: Optimization Problems

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Basic Calculus – Grade 11 Alternative Delivery Mode Quarter 3 – Module 9: Optimization Problems First Edition, 2020 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

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Basic Calculus Quarter 3 – Module 9: Optimization Problems

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Introductory Message This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson. Each SLM is composed of different parts. Each part shall guide you step-bystep as you discover and understand the lesson prepared for you. Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these. 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.

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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.

When you finish this module, you will be able to: 1. solve optimization problems that yield polynomial functions. (STEM_BC11D-IIIg-1)

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What I Know

Answer the following problems on a separate sheet of paper. 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 7. Calculate the dimensions of a rectangle whose perimeter is 150 m and provide a maximum area. A. 50𝑚 𝑎𝑛𝑑 10𝑚 C. 35𝑚 𝑎𝑛𝑑 25𝑚 B. 37.5𝑚 𝑎𝑛𝑑 37.5𝑚 D. 30.5𝑚 𝑎𝑛𝑑 30.5𝑚 8. Rex inherits a piece of land. He wants to install fences on its rectangular region and maximize the a rea. 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 10. A carton board with 8 cm by 8 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. A. 𝑙 𝑎𝑛𝑑 𝑤 =

B. 𝑙 𝑎𝑛𝑑 𝑤 = C. 𝑙 𝑎𝑛𝑑 𝑤

D. 𝑙 𝑎𝑛𝑑 𝑤

16 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 3 14

= 4/3𝑐𝑚

𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 = 4/5𝑐𝑚

3 17 = 3 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 = 16 = 5 𝑐𝑚, ℎ𝑒𝑖𝑔ℎ𝑡 =

5/3𝑐𝑚

2/3𝑐𝑚

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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

Answer the following items on a separate sheet of paper. 1. What is the area of a rectangle if its dimensions are 5m by 7m? 2. Calculate the perimeter of a square with 4 meters per side. 3. Given two equations 3𝑥 − 𝑦 = 2 and 𝑥 + 1 = 𝑦. Find the values of 𝑥 and 𝑦. 4. Find the derivative of 𝑦 = 𝑥 2 + 7𝑥 − 3.

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. A rectangular basketball court has a dimension of 20m by 10m. Suddenly, a man came and asked for the area of the basketball court. What will be your answer and detailed solution for his question?

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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.

•

Constraint is a conditional concept that can be transformed into an equation which is part of an optimization problem. Most problems have a given constant quantity.

•

Optimization equation is part of the problem that needs to be maximized or minimized.

Below are guides in solving optimization problems. 1. Read and understand the problem carefully. 2. Create a sketch or illustration with given labels whenever necessary. 3. List what is asked and what are the given data in the problem. 4. Write down the constraint and optimization equations. 5. Equate one variable in terms of the other from the constraint equation. Then replace or substitute it to the same variable on the optimization equation. 6. Simplify the optimization equation then solve for its first derivative. 7. Set the first derivative equation to zero and find the critical point. 8. Test the critical point values using the derivative tests. 9. Substitute the critical value to solve for the other.

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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.

1. List what is asked on the problem. 2. Write the constraint and the optimization equation.

𝑥 ∙ 𝑦 = 121

𝑀 =𝑥+𝑦

Constraint equation

Optimization equation (minimized)

121 𝑦= 𝑥

3. On the constraint equation, solve for 𝑦. Then, substitute that equation to its corresponding 𝑦 on the optimization equation.

𝑀=𝑥+(

121 ) 𝑥

𝑀 = 𝑥 + 121𝑥

−1

𝑀 ′ = 1 + (−1)(121𝑥 −2)

4. Simplify and take its first derivative.

𝑀′ = 1 −

121 𝑥2

;

121 𝑥2 121 121 121 0+ 2 =1− 2 + 2 𝑥 𝑥 𝑥 121 =1 𝑥2 121 = 𝑥2 𝑥 = +√121 𝒙 = +𝟏𝟏 0=1−

5. Set the equation to zero and solve for the 𝑥 value (critical point). Hint: use algebraic manipulation to solve for x.

6. Test the 𝑥 value. Substitute it to the second derivative and check whether the answer is less than or greater than zero. Since the answer is greater than zero then it is the minimum. value.

Test:

121 𝑥2 𝑀 ′′ = 0 − (−2)121𝑥 −3 𝑀′ = 1 −

𝑀 ′′ =

242 𝑥3

242 ≈ 0.18 (11)3 𝑀 ′′ > 0

𝑀 ′′ =

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7. Substitute the minimum 𝑥 value to the simplified constraint equation to solve for 𝑦.

If 𝑥 = 1, then

𝑦=

121 = +11 11

Thus, 𝑥 = 11 and 𝑦 = 11 are the two positive numbers. Example 2 Solve for the dimensions of a rectangle whose perimeter is 250 m with a maximum area. Solution: Let 𝑥 and 𝑦 be the missing sides of the rectangle with perimeter 𝑃 = 250𝑚 .

𝑥 1. Draw a diagram. List what is asked on the problem and label the diagram with relevant data.

𝑦

𝑦

𝑥

2𝑥 + 2𝑦 = 250

2. Write the constraint and the optimization equation.

Constraint equation 3. On the constraint equation, solve for 𝑦. Then, substitute that equation to its corresponding 𝑦 on the optimization equation.

𝑦 = 125 − 𝑥

𝐴= 𝑥∙𝑦 Optimization equation (maximized)

𝐴 = 𝑥 ∙ (125 − 𝑥 ) 𝐴 = 125𝑥 − 𝑥 2 𝐴′ = 125 − 2𝑥

4. Simplify and take its first derivative.

0 = 125 − 2𝑥 125 𝑥= 2 𝑥 = 𝟔𝟐. 𝟓 𝒎

5. Set the equation to zero and solve for the 𝑥 value (critical point).

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6. Test the 𝑥 value. Substitute it to the second derivative and check whether the answer is less than or greater than zero. Since the answer is less than zero, then it is the maximum value.

Test:

𝐴′ = 125 − 2𝑥

𝐴′′ = −2

𝐴′′ < 0

If 𝑥 = 62.5 𝑚, then

7. Substitute the maximum 𝑥 value to the simplified constraint equation to solve for 𝑦.

𝑦 = 125 − (62.5) = 62.5𝑚

Thus, 𝑥 = 62.5 𝑚 by 𝑦 = 62.5 𝑚 are the dimensions of the rectangle.

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 optimization equation.

Constraint equations

𝑙 = 12𝑥 − 2𝑥 𝑤 = 12𝑥 − 2𝑥 ℎ=𝑥

𝑠𝑖𝑑𝑒𝑠 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 [0, 6] Optimization equation (maximized)

𝑉 =𝑙∙𝑤∙ℎ

3. From the constraint equations, substitute them to its corresponding 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑤𝑖𝑑𝑡ℎ and ℎ𝑒𝑖𝑔ℎ𝑡 on the optimization equation.

𝑉 = (12 − 2𝑥)(12 − 2𝑥 )(𝑥) 𝑉 = 4𝑥 3 − 48𝑥 2 + 144𝑥

4. Simplify and take its first derivative.

𝑉 ′ = 12𝑥 2 − 96𝑥 + 144

12𝑥2 − 96𝑥 + 144 = 0

5. Set the equation to zero and solve for the 𝑥 value (critical point).

(6𝑥 − 12)(2𝑥 − 12) = 0 6𝑥 − 12 = 0 6𝑥 12 = 6 6 𝑥=𝟐

6. Since there are two critical values, we have to check which one will give a sensible answer by substituting them to the volume equation. The result suggests that 𝑥 = 2 is the correct value.

If

and

2𝑥 − 12 = 0 2𝑥 12 = 2 2 𝑥=𝟔

𝑥 = 2, then

𝑉 = [12 − 2(2)][12 − 2(2)][2] = 128 𝑐𝑚3  If 𝑥

= 6, then

𝑉 = [12 − 2(6)][12 − 2(6)][6] = 0 𝑐𝑚3 x

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7. Test the 𝑥 value. Substitute it to the second derivative and check whether the answer is less than or greater than zero. Since the answer is less than zero, then it is the maximum value.

Test:

8. Substitute the maximum 𝑥 value to the simplified constraint equation to solve for 𝑙 and 𝑤.

Solving for the dimensions:

𝑉 ′ = 12𝑥 2 − 96𝑥 + 144 𝑉 ′′ = 24𝑥 − 96 𝑉 ′′ = 24(2) − 96 = −48 𝑉 ′′ < 0

ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑥 = 2 𝑐𝑚

𝑙𝑒𝑛𝑔𝑡ℎ = 12 − 2(𝑥) = 12 − 2(2) = 8 𝑐𝑚

𝑤𝑖𝑑𝑡ℎ = 12 − 2(𝑥) = 12 − 2(2) = 8 𝑐𝑚

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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.

Solution

Steps

Constraint equation: 1. List what is asked on the problem. 2. Write the constraint optimization equations.

and

the Optimization equation:

3. On the constraint equation, solve for 𝑦. Then, substitute that equation to its corresponding 𝑦 on the optimization equation.

4. Simplify and take its first derivative.

5. Set the equation to zero and solve for the 𝑥 value (critical point). 6. Test the 𝑥 value. Substitute it to the second derivative and check whether the answer is less than or greater than zero. Since the answer is greater than zero, then it is the minimum. 7. Substitute the minimum 𝑥 value to 𝑥 = _______ and 𝑦 = _______ are the simplified constraint equation to positive numbers solve for 𝑦.

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the

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2. Determine the dimensions of a rectangle whose perimeter is 300 m with a maximum area.

Solution

Steps 1. Draw a diagram. List what is asked on the problem and label the diagram with relevant data.

Constraint equation: 2. Write the constraint optimization equations.

and

the Optimization equation:

3. On the constraint equation, solve for 𝑦. Then, substitute that equation to its corresponding 𝑦 on the optimization equation.

4. Simplify and take its first derivative.

5. Set the equation to zero and solve for the 𝑥 value (critical point).

6. Test the 𝑥 value. Substitute it to the second derivative and check whether the answer is less than or greater than zero.

7. Substitute the minimum 𝑥 value to 𝑥 = _______ and 𝑦 = _______ are the simplified constraint equation to dimensions of the rectangle. solve for 𝑦.

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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.

Solution

Steps 1. Draw a diagram. List what is asked on the problem and label the diagram with relevant data.

Constraint equation: 2. Write the constraint optimization equations.

and

the Optimization equation:

3. Substitute the constraint equations, to the corresponding 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑤𝑖𝑑𝑡ℎ and ℎ𝑒𝑖𝑔ℎ𝑡 of the optimization equation. 4. Simplify and take its first derivative.

5. Set the equation to zero and solve for the 𝑥 value (critical point). 6. If there are two critical values, we have to check which one will give a sensible answer by substituting them to the volume equation. 7. Test the 𝑥 value. Substitute it to the second derivative and check whether the answer is less than or greater than zero. 8. Substitute the maximum 𝑥 value to the simplified constraint equation to solve for 𝑙 and 𝑤.

9. Solve for the dimensions.

ℎ𝑒𝑖𝑔ℎ𝑡 = ________ 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. 1. Describe an optimization problem.

2. How can we solve optimization problems?

3. Mention past topics in mathematics that you applied on the last activity to solve optimization problems.

What I Can Do

Answer the word problem on a separate sheet of paper. 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

A. 𝑎 = , 𝑏 = 3

√3 2

1

B. 𝑎 = , 𝑏 = 2

√2 2

1

C. 𝑎 = , 𝑏 = 4

√5 2

1

D. 𝑎 = , 𝑏 = 6

√7 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 ....