Applied max and min. Georgia owns a piece of land along the Ogeechee River She wants to fence in her garden using the river as one side.

Slides:



Advertisements
Similar presentations
The Derivative in Graphing and Applications
Advertisements

3.7 Modeling and Optimization
OPTIMIZATION © Alex Teshon Daffy Durairaj.
To optimize something means to maximize or minimize some aspect of it… Strategy for Solving Max-Min Problems 1. Understand the Problem. Read the problem.
Optimization 4.7. A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that.
Applications of Differentiation
4.4 Optimization Finding Optimum Values. A Classic Problem You have 40 feet of fence to enclose a rectangular garden. What is the maximum area that you.
Modeling and Optimization
4.5 Optimization Problems Steps in solving Optimization Problems 1.Understand the Problem Ask yourself: What is unknown? What are the given quantities?
APPLICATIONS OF DIFFERENTIATION 4. The methods we have learned in this chapter for finding extreme values have practical applications in many areas of.
4.6 Optimization The goal is to maximize or minimize a given quantity subject to a constraint. Must identify the quantity to be optimized – along with.
Section 3.7 – Optimization Problems. Optimization Procedure 1.Draw a figure (if appropriate) and label all quantities relevant to the problem. 2.Focus.
Applications of Differentiation Section 4.7 Optimization Problems
Lesson 4.4 Modeling and Optimization What you’ll learn about Using derivatives for practical applications in finding the maximum and minimum values in.
AIM: APPLICATIONS OF FUNCTIONS? HW P. 27 #74, 76, 77, Functions Worksheet #1-3 1 Expressing a quantity as a function of another quantity. Do Now: Express.
Applied Max and Min Problems Objective: To use the methods of this chapter to solve applied optimization problems.
4.7 Applied Optimization Wed Jan 14
CHAPTER 3 SECTION 3.7 OPTIMIZATION PROBLEMS. Applying Our Concepts We know about max and min … Now how can we use those principles?
Applied Max and Min Problems
Section 4.4: Modeling and Optimization
{ ln x for 0 < x < 2 x2 ln 2 for 2 < x < 4 If f(x) =
Linearization , Related Rates, and Optimization
Section 4.4 Optimization and Modeling
Lesson 4-7 Optimization Problems. Ice Breaker Using a blank piece of paper: Find the extrema (maximum) of A(w) = 2400w – 2w² A’(w) = 2400 – 4w A’(w) =
1.Read 3. Identify the known quantities and the unknowns. Use a variable. 2.Identify the quantity to be optimized. Write a model for this quantity. Use.
4.4 Modeling and Optimization Buffalo Bill’s Ranch, North Platte, Nebraska Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly,
4.7 Optimization Problems In this section, we will learn: How to solve problems involving maximization and minimization of factors. APPLICATIONS OF DIFFERENTIATION.
Optimization. Objective  To solve applications of optimization problems  TS: Making decisions after reflection and review.
OPTIMIZATION.
Warmup- no calculator 1) 2). 4.4: Modeling and Optimization.
Section 4.5 Optimization and Modeling. Steps in Solving Optimization Problems 1.Understand the problem: The first step is to read the problem carefully.
4.1 Extreme Values of Functions
Applied Max and Min Problems (Optimization) 5.5. Procedures for Solving Applied Max and Min Problems 1.Draw and Label a Picture 2.Find a formula for the.
Optimization Problems
Optimization. First Derivative Test Method for finding maximum and minimum points on a function has many practical applications called Optimization -
Applied max and min. 12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x)
Optimization Problems Section 4-4. Example  What is the maximum area of a rectangle with a fixed perimeter of 880 cm? In this instance we want to optimize.
2.7 Mathematical Models. Optimization Problems 1)Solve the constraint for one of the variables 2)Substitute for the variable in the objective Function.
2.7 Mathematical Models Some will win, some will lose, some are born to sing the blues. Oh the movie never ends, it goes on and on and on and on. -Journey.
A25 & 26-Optimization (max & min problems). Guidelines for Solving Applied Minimum and Maximum Problems 1.Identify all given quantities and quantities.
4.4 Modeling and Optimization, p. 219 AP Calculus AB/BC.
Sec 4.6: Applied Optimization EXAMPLE 1 An open-top box is to be made by cutting small congruent squares from the corners of a 12-in.-by-12-in. sheet of.
Applied max and min. Steps for solving an optimization problem Read the problem drawing a picture as you read Label all constants and variables as you.
6.2: Applications of Extreme Values Objective: To use the derivative and extreme values to solve optimization problems.
Sec 4.7: Optimization Problems EXAMPLE 1 An open-top box is to be made by cutting small congruent squares from the corners of a 12-in.-by-12-in. sheet.
STEPS IN SOLVING OPTIMIZATION PROBLEMS 1.Understand the Problem The first step is to read the problem carefully until it is clearly understood. Ask yourself:
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Maximum-Minimum (Optimization) Problems OBJECTIVE  Solve maximum and minimum.
Ch. 5 – Applications of Derivatives 5.4 – Modeling and Optimization.
Optimization Problems
Chapter 12 Graphs and the Derivative Abbas Masum.
Sect. 3-7 Optimization.
4.5 Optimization II Dr. Julia Arnold
Ch. 5 – Applications of Derivatives
5-4 Day 1 modeling & optimization
AIM: How do we use derivatives to solve Optimization problems?
4.5 Optimization II Dr. Julia Arnold
Honors Calculus 4.8. Optimization.
Applied Max and Min Problems
Calculus I (MAT 145) Dr. Day Wednesday Nov 8, 2017
Calculus I (MAT 145) Dr. Day Friday Nov 10, 2017
Optimization Chapter 4.4.
7. Optimization.
4.6 Optimization The goal is to maximize or minimize a given quantity subject to a constraint. Must identify the quantity to be optimized – along with.
Optimization Problems
Optimization Problems
Sec 4.7: Optimization Problems
Calculus I (MAT 145) Dr. Day Wednesday March 27, 2019
Calculus I (MAT 145) Dr. Day Monday April 8, 2019
Presentation transcript:

Applied max and min

Georgia owns a piece of land along the Ogeechee River She wants to fence in her garden using the river as one side.

She also owns 1000 ft of fence to make the rectangular garden She wants to fence in her garden using the river as one side. F = F = F =

She owns 1000 ft of fence Write a secondary equation Write a secondary equation Usually the first thing given F = x +

She owns 1000 ft of fence Write a secondary equation Write a secondary equation Usually the first thing given F = x + y + x 1000 = 2x + y is the secondary

She owns 1000 ft of fence Write a secondary equation Write a secondary equation Usually the first thing given F = x + y + x 1000 = 2x + y is the secondary Solve for y y = 1000 – 2x

What is the area of a possible garden? A = L * W A= 5 * 990 A = 4950 sq. ft.

What is the largest possible area? Find the variable that you want to optimize and write the primary equation Find the variable that you want to optimize and write the primary equation

What is the area of the shown garden? A. A. A = 2xy sq. feet B. B. A = 2x + y feet C. C. A = xy sq. feet

What is the area of the largest possible garden? A = x * y primary

Place y into the primary y = 1000 – 2x (secondary) A = x * y (primary) A = x * (1000 – 2x) A = 1000x – 2x 2

A = 1000x – 2x 2 If A’ = 0, find x. A. A. x = 250 feet B. B. x = 300 feet C. C. x = 350 feet

Differentiate the primary and set to zero A = 1000x – 2x 2 A’ = 1000 – 4x = = 4x 1000 = 4x 250 = x 250 = x

What is the area of the largest possible garden? A’ = 1000 – 4x = 0 A’’ = -4 concave down A’’(250) = -4 Relative max at x = 250 A = 250 * 500 = 125,000 sq. ft.

Girth is the smaller distance around the object

Post office says the max Length + girth is 108 A. A. 108 = L + x B. B. 108 = L + 2x C. C. 108 = L + 4x

Find x that maximizes the volume A. A. V = 4x + L B. B. V = x 2 * L C. C. V = 4x * L

V = x 2 * L L = 108 – 4x V = V = x 2 * (108 – 4x) = 108 x x 3 V’ = 216 x – 12 x 2 = 0 12x(18 – x) = 0 x = 18 V’’ = 216 – 24x and if x = 18, V’’ is Negative => local max.

If the volume is 357 cm 3 Minimize the aluminum. Minimize the aluminum.

V = A. V =  r 2 B. V = hr 3 C. V =  r 2 h

If the volume is 357 cc 3 Minimize the aluminum. Minimize the aluminum. V =  r 2 h = 357 V =  r 2 h = 357 h = 357/(  r 2 ) h = 357/(  r 2 ) A =  r 2 + A =  r 2 +  r 2 +  r  rh 2  rh

A = A. A =  r 2 +2  rh B. A = 2  r 2 +2  rh C. A =  r 2 h+2  h

h = 357/(  r 2 ) A =  r 2 +  r  rh Minimize the aluminum. Minimize the aluminum. A = 2  r  r 357 /(  r 2 ) A = 2  r  r 357 /(  r 2 ) = 2  r r -1 = 2  r r -1 A’ = 4  r - 714r -2 =0 A’ = 4  r - 714r -2 =0 4  r 3 =  r 3 = 714

4  r 3 = 714 r 3 = r 3 = 714/(4  714/(4  r = r = Diameter = cm = in Diameter = cm = in

12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x)

12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x) V’=x(12 - 2x)(-2)+

12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x) V’=x(12 - 2x)(-2)+x(-2) (12 - 2x)+

12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x) V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)

12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x) V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)=(12-2x)(-2x-2x+(12-2x))

12” by 12” sheet of cardboard Find the box with the most volume. V = x(12 - 2x)(12 - 2x) V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)=(12-2x)(-2x-2x+(12-2x)) =(12-2x)(12-6x)=144-24x-72x+12x 2 =12(12-8x+x 2 )

12” by 12” sheet of cardboard Find the box with the most V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)=(12-2x)(-2x-2x+(12-2x)) =(12-2x)(12-6x)=144-24x-72x+12x 2 =12(12-8x+x 2 ) = 12(x-6)(x-2) = 0

12” by 12” sheet of cardboard Find the box with the most V’ = 12(12-8x+x 2 ) = 12(x-6)(x-2) = 0 x = 2 or x = 6

V = (12-2x)(12-2x) x = 144x - 48x 2 + 4x 3 Find the box with the most volume. dV/dx = 144 – 96 x + 12 x 2 = 0 when 12(12 - 8x+ x 2 ) = 0 (6 – x)(2 – x) = 0

dV/dx = 144 – 96 x + 12 x 2 X = 2 or x = 6 d 2 V/dx 2 = x At x = 2 or at x = 6 negative positive

dV/dx = 144 – 96 x + 12 x 2 X = 2 or x = 6 d 2 V/dx 2 = x At x = 2 or at x = 6 negative positive Local max local min

Steps for solving an optimization problem Read the problem drawing a picture as you read Label all constants and variables as you read If you have two unknowns, write a secondary equation Usually the first thing given Find the variable that you want to optimize and write the primary equation Eliminate one variable from the primary equation using the secondary equation Determine the domain of the new primary equation Differentiate the primary equation Set the derivative equal to zero Solve for the unknown Check the endpoints or run a first or second derivative test

Read the problem drawing a picture as you read Label all constants and variables as you read Inside a semicircle of radius R.

Semicircle of radius 6. If you have two unknowns, write a secondary equation. Usually the first thing given.

Write the equation of a circle, centered at the origin of radius 6. A. x + y = 36 B. x 2 + y 2 = 6 C. x 2 + y 2 = 36 D. y =

We identify the primary equation by the key word maximizes or minimizes Find the value of x that maximizes the blue area.

Find the rectangle with the largest area Find the value of x that maximizes the blue area.

Which of the following is the primary equation? A. A = x y B. A = 2 x y C. A = ½ x y D. A = 4 x y

Eliminate one variable from the primary equation using the secondary equation A(x) = 2xy = 2x(6 2 - x 2 ) ½ A 2 = 4x 2 (36 - x 2 ) = 144x 2 - 4x 4

Differentiate A 2 = 144x 2 - 4x 4 implicitly. A. A’ = 288x - 16x 3 B. 2AA’ = 144x - 8x C. A’ = 144x – 16x D. 2AA' = 288x - 16x 3

AA' = 144x - 8x 3 = 0 Solve for x AA' = 144x - 8x 3 = 0 Solve for x A. x= 0, 3 root(2), - 3 root(2) B. x = 6 root(2), - 6 root(2) C. x = 0, 3, -3 D. x = 3/root(2), - 3/root(2)

Check the endpoints or run a first or second derivative test AA' = 18x - x 3 = x(18 – x 2 ) A' = 0 when x = 3 root(2) or x = 0 AA’(3)= > 0 AA’(6) = < 0

Check the endpoints or run a first or second derivative test AA' = 18x - x 3 = x(18 – x 2 ) A' = 0 when x = 3 root(2) or x = 0 AA’(3)= > 0 AA’(6) = < 0 AA”+A’A’ = 18 – 3 x 2 =0 when ??

Check the endpoints or run a first or second derivative test AA' = 18x - x 3 = x(18 – x 2 ) A' = 0 when x = 3 root(2) or x = 0 AA’(3)= > 0 AA’(6) = < 0 AA”+A’A’ = 18 – 3 x 2 =0 when AA” = (18 – 3x 2 ) – x 2 (18 – x 2 ) 2

AA’(3)= > 0 AA’(6) = AA’(6) = < 0 A. There is a local max at x = 3 root(2) B. Neither a max nor min at 3 root(2) C. There is a local min at x = 3 root(2) D. Inflection point at x = 3 root(2)

Steps for solving an optimization problem Read the problem drawing a picture as you read Label all constants and variables as you read If you have two unknowns, write a secondary equation Usually the first thing given Find the variable that you want to optimize and write the primary equation Eliminate one variable from the primary equation using the secondary equation Determine the domain of the new primary equation Differentiate the primary equation Set the derivative equal to zero Solve for the unknown Check the endpoints or run a first or second derivative test

Build a rain gutter with the dimensions shown. Base Area = h(b+1) BA = sin(  )[cos(  )+1] V=

BA = sin(  )[cos(  )+1] V= A. 20 sin(  )[cos(  )+1] B. 20 sin(  ) 2 [cos(  )+1] 2 C. sin(  )[cos(  )+1] 2 D. sin(  ) 2 [cos(  )+1]

Find  that maximizes the volume V = 20 BA V = 20 sin(  )[cos(  ) + 1 ] V’ =

V=20 sin(  )[cos(  )+1] dV/d  = A. 20 sin(  )[cos(  )+1] B. 20 cos(  ) sin(  ) C. 20[sin(  )(- sin(  ))+(cos(  )+1)cos(  )] D cos(  ) sin(  )

Find  that maximizes the volume V’ = 20sin(  )[-sin(  )]+[cos(  ) + 1]20 cos(  ) = 20 cos 2 (  ) - 20 sin 2 (  ) + 20 cos(  ) = 20[2 cos(  ) - 1][cos(  ) + 1] = 0

20[2 cos(  ) - 1][cos(  ) + 1] = 0 Solve for  on . A.  /3 or  B.  /2 or  C.  /6 or  D.  /3 or 

Find  that maximizes the volume of the gutter V’ = 20[2 cos(  ) - 1][cos(  ) + 1] = 0 2 cos(  ) = 1 or cos(  ) = -1  or 

Find  that maximizes the volume of the gutter V’ = 20 cos 2 (  ) - 20 sin 2 (  ) + 20 cos(  ) V’’ = -40 cos(  )sin(  ) – 40 sin(  )cos(  ) -20 sin(  ) V’’(  ) = -40(½) – 40 (½) - 20 a local maximum at x =  a local maximum at x =  V”(  ) = 0 -> Test fails

Local max at  =  /3 V’ = 20 cos 2 (  ) - 20 sin 2 (  ) + 20 cos(  ) V’(  /2) = -20 V’(3  /2) = -20 Second derivative test failed First derivative test says decreasing on [  /3,  ]

GSU builds 400 meter track. 400 = 2x +  d

Soccer requires a maximum of green area A = xd, but d = because 400 = 2x +  d So A =

Soccer requires a maximum green rectangle So A = and A’ = when x = 100 meters and A’ = when x = 100 meters A” =

Soccer requires a maximum green area 400 = 2x +  d and when x = 100 meters 200 =  d or d = 200 / 