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3.7: Optimization Homework: p , 19, 21, 29, 33, 47

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Presentation on theme: "3.7: Optimization Homework: p , 19, 21, 29, 33, 47"β€” Presentation transcript:

1 3.7: Optimization Homework: p.220 17, 19, 21, 29, 33, 47
Standards EK2.3C3 – The derivative can be used to solve optimization problems, that is, finding a maximum or minimum value of a function over a given interval Learning Objectives: Solve applied minimum and maximum problems

2 Does 𝑽= 𝒔 πŸ‘ or does 𝑽= 𝒔 𝟐 ⋅𝒉 or does 𝑽=π’β‹…π’˜β‹…π’‰?
Concept 1: Formulas Does 𝑽= 𝒔 πŸ‘ or does 𝑽= 𝒔 𝟐 ⋅𝒉 or does 𝑽=π’β‹…π’˜β‹…π’‰?

3 The Equation/Formula that is to be used in the optimization
Concept 1: Vocabulary Primary Equation The Equation/Formula that is to be used in the optimization

4 Teacher Example 1: Finding Maximum Volume
An open top box with a square base needs to have a surface area of 108 cubic inches. How large would the sides have to be to maximize the volume?

5 Teacher Example 1: Finding Maximum Volume
Step 1: Volume Formula Step 2: Surface Area Step 3: Single Variable Conversion Step 4: Feasible Region Step 5: Maximize Step 6: Solve for the variable Step 7: Evaluate

6 Teacher Example 1: Finding Maximum Volume

7 Concept 1 (Continued): Guidelines for Solving Optimization Problems
On page 216, bottom of the page, enter the appropriate information into your notes. 2 minutes + 10 minutes for SLE

8 Student Led Example 1: Finding Maximum Volume
Handout: Problem 1

9 Student Led Example 1: Handout – Problem 1
This image is here to ASSIST and has nothing to do with the actual problem…they are, however, ridiculously similar.

10 Student Led Example 1: Finding Maximum Volume

11 Student Led Example 1: Finding Maximum Volume
𝑽= πŸ‘πŸŽβˆ’πŸπ’™ πŸπŸ’βˆ’πŸπ’™ ⋅𝒙 𝑽 β€² =𝟏𝟐 𝒙 𝟐 βˆ’πŸπŸ•πŸ”π’™+πŸ’πŸπŸŽ =πŸ’(πŸ‘π’™βˆ’πŸ‘πŸ“)(π’™βˆ’πŸ‘) 10 Minutes

12 Student Led Example 1: Finding Maximum Volume
πŸŽβ‰€π’™β‰€πŸ• Feasible Region: Critical Points: 𝒙=πŸ‘

13 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ?

14 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ?

15 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ? What quantity needs to be minimized? Or, in other words, what is the PRIMARY EQUATION?

16 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ? 𝒅= π’™βˆ’πŸŽ 𝟐 + π’šβˆ’πŸ 𝟐 π’š=πŸ’βˆ’ 𝒙 𝟐 𝒅= 𝒙 𝟐 + πŸ’βˆ’ 𝒙 𝟐 βˆ’πŸ 𝟐 𝒅= 𝒙 πŸ’ βˆ’πŸ‘ 𝒙 𝟐 +πŸ’

17 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ? 𝒅 is smallest when the radicand is smallest. 𝒙 πŸ’ βˆ’πŸ‘ 𝒙 𝟐 +πŸ’ β€² =πŸ’ 𝒙 πŸ‘ βˆ’πŸ”π’™ =πŸπ’™ 𝒙 𝟐 βˆ’πŸ‘ Critical values occur at 𝒙= 𝟎,Β± πŸ‘ 𝟐

18 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ?

19 Teacher Example 2: Minimum Distance
Which points on the graph π’š=πŸ’βˆ’ 𝒙 𝟐 are closest to the point 𝟎,𝟐 ? 𝒇 πŸ‘ 𝟐 = πŸ“ 𝟐 𝒇 βˆ’ πŸ‘ 𝟐 = πŸ“ 𝟐

20 Student Led Example 2: Finding Minimum Distance
Handout: Problem 2

21 Student Led Example 2: Finding Minimum Distance
Function to be minimized: 𝒅= π’™βˆ’πŸ• 𝟐 + 𝒙 𝟐 Critical Points: 𝒙= πŸπŸ‘ 𝟐 Feasible Region: βˆ’βˆž<𝒙<∞

22 Teacher Example 3: Finding Minimum Length
Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?

23 Teacher Example 3: Finding Minimum Length
Pictures! 

24 Teacher Example 3: Finding Minimum Length
Function to be Minimized: 𝑾=π’š+𝒛

25 Teacher Example 3: Finding Minimum Length
Function to be Minimized: 𝑾=π’š+𝒛 π’š= 𝒙 𝟐 +πŸπŸ’πŸ’ 𝒛= πŸ‘πŸŽβˆ’π’™ 𝟐 + 𝟐 πŸ– 𝟐 = 𝒙 𝟐 βˆ’πŸ”πŸŽπ’™+πŸπŸ”πŸ–πŸ’

26 Teacher Example 3: Finding Minimum Length
Function to be Minimized: 𝑾=π’š+𝒛 π‘Š β€² = 𝑦 β€² +𝑧′ 𝑧 β€² = π‘₯βˆ’30 π‘₯ 2 βˆ’60π‘₯+1684 𝑦 β€² = π‘₯ π‘₯ π‘‘π‘Š 𝑑π‘₯ = π‘₯ π‘₯ π‘₯βˆ’30 π‘₯ 2 βˆ’60π‘₯+1684

27 Teacher Example 3: Finding Minimum Length
Function to be Minimized: 𝑾=π’š+𝒛 π‘₯ π‘₯ π‘₯βˆ’30 π‘₯ 2 βˆ’60π‘₯ =0 After some ridiculous algebra we get… π‘₯={9,βˆ’22.5}

28 Teacher Example 3: Finding Minimum Length
Function to be Minimized: 𝑾=π’š+𝒛 π‘₯= 9,βˆ’22.5 π‘Š π‘₯ =𝑦 π‘₯ +𝑧 π‘₯ 𝑦 9 = = = =15 𝑧 π‘₯ = βˆ’ = 81βˆ’ = =35 π‘Š 9 =50

29 Student Led Example 3: Finding Minimum Length
Handout: Problem 3

30 Exit Task – Complete if not done
Enter into your notes: β€œGuidelines for Solving Applied Minimum and Maximum problems” on page 216


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