Applied max and minApplied max and min

Georgia owns a piece of Georgia owns a piece of land along the Ogeechee land along the Ogeechee RiverRiverShe wants to fence in her garden She wants to fence in her garden

using the river as one side.using the river as one side.

She also owns 1000 ft of She also owns 1000 ft of fence to make the fence to make the rectangular gardenrectangular garden

She wants to fence in her garden She wants to fence in her garden using the river as one side.using the river as one side.

F = 1 + 998 + 1F = 1 + 998 + 1

F = 5 + 990 + 5F = 5 + 990 + 5

F = 20 + 960 + 20F = 20 + 960 + 20

She owns 1000 ft of She owns 1000 ft of fencefence

Write a secondary equation Write a secondary equation Usually the first thing givenUsually the first thing given

F = x +F = x +

She owns 1000 ft of She owns 1000 ft of fencefence

Write a secondary equation Write a secondary equation Usually the first thing givenUsually the first thing given

F = x + y + xF = x + y + x

1000 = 2x + y is the secondary1000 = 2x + y is the secondary

She owns 1000 ft of She owns 1000 ft of fencefence

Write a secondary equation Write a secondary equation Usually the first thing givenUsually the first thing given

F = x + y + xF = x + y + x

1000 = 2x + y is the secondary1000 = 2x + y is the secondary

Solve for ySolve for y

y = 1000 – 2xy = 1000 – 2x

What is the area of a What is the area of a possible garden?possible garden?

A = L * WA = L * W

A= 5 * 990A= 5 * 990

A = 4950 sq. ft.A = 4950 sq. ft.

What is the largest What is the largest possible area?possible area?

Find the variable that you want to Find the variable that you want to optimize and write the primary optimize and write the primary equation equation

What is the area of the What is the area of the shown garden?shown garden?

A. A = 2xy sq. feetB. A = 2x + y feetC. A = xy sq. feet

What is the area of the What is the area of the largest possible garden?largest possible garden?

A = x * y primaryA = x * y primary

Place y into the primaryPlace y into the primary

y = 1000 – 2x (secondary)y = 1000 – 2x (secondary)

A = x * y (primary)A = x * y (primary)

A = x * (1000 – 2x)A = x * (1000 – 2x)

A = 1000x – 2xA = 1000x – 2x22

A = 1000x – 2xA = 1000x – 2x22 If A’ = 0, find x.If A’ = 0, find x.

A. x = 250 feetB. x = 300 feetC. x = 350 feet

Differentiate the primaryDifferentiate the primaryand set to zeroand set to zero

A = 1000x – 2xA = 1000x – 2x22

A’ = 1000 – 4x = 0A’ = 1000 – 4x = 0

1000 = 4x1000 = 4x

250 = x250 = x

What is the area of the What is the area of the largest possible garden?largest possible garden?

A’ = 1000 – 4x = 0A’ = 1000 – 4x = 0

A’’ = -4 concave downA’’ = -4 concave down

A’’(250) = -4A’’(250) = -4

Relative max at x = 250Relative max at x = 250

A = 250 * 500 = 125,000 sq. ft.A = 250 * 500 = 125,000 sq. ft.

Girth is the smaller distance around the object

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

Find x that maximizes the volume

A. V = 4x + LB. V = x2 * LC. V = 4x * L

V = x2 * LL = 108 – 4x

V = V = x2 * (108 – 4x) = 108 x2 - 4 x3 V’ = 216 x – 12 x2 = 012x(18 – x) = 0x = 18V’’ = 216 – 24x and if x = 18, V’’ isNegative => local max.

If the volume is 357 If the volume is 357 cmcm33

Minimize the aluminum.Minimize the aluminum.

V = V =

A.A. V = V = rr22

B.B. V = hrV = hr33

C.C. V = V = rr22hh

If the volume is 357 ccIf the volume is 357 cc33

Minimize the aluminum.Minimize the aluminum. V = V = rr22h = 357h = 357 h = 357/(h = 357/(rr22)) A = A = rr2 2 ++

rr2 2 ++

22rhrh

A = A =

A.A. A = A = rr22+2+2rhrh

B.B. A = 2A = 2rr22+2+2rhrh

C.C. A = A = rr22h+2h+2hh

h = 357/(h = 357/(rr22))A = A = rr2 2 ++ rr2 2 ++ 22rhrh

Minimize the aluminum.Minimize the aluminum. A = 2A = 2 rr2 2 ++ 22r 357 /(r 357 /(rr22)) = 2= 2 rr2 2 ++ 714r714r-1-1 A’ = 4A’ = 4r - 714rr - 714r-2-2=0=0 44rr3 3 = 714 = 714

2

7144 r

r

44 r r3 3 = 714 = 714

rr3 3 = = 714/(4714/(4 r =r =

Diameter = 7.6888 cm = 3.0271 Diameter = 7.6888 cm = 3.0271 in in

3714

4

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most volume.volume.

V = V =

x(12 - 2x)(12 - 2x)x(12 - 2x)(12 - 2x)

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

volume.volume.

V = x(12 - 2x)(12 - 2x)V = x(12 - 2x)(12 - 2x)

V’=x(12 - 2x)(-2)+V’=x(12 - 2x)(-2)+

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

volume.volume.

V = x(12 - 2x)(12 - 2x)V = x(12 - 2x)(12 - 2x)

V’=x(12 - 2x)(-2)+x(-2) (12 - 2x)+ V’=x(12 - 2x)(-2)+x(-2) (12 - 2x)+

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

volume.volume.

V = x(12 - 2x)(12 - 2x)V = x(12 - 2x)(12 - 2x)

V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)2x)

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

volume.volume.

V = x(12 - 2x)(12 - 2x)V = x(12 - 2x)(12 - 2x)

V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)2x)

=(12-2x)(-2x-2x+(12-2x))=(12-2x)(-2x-2x+(12-2x))

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

volume.volume.

V = x(12 - 2x)(12 - 2x)V = x(12 - 2x)(12 - 2x)

V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)2x)

=(12-2x)(-2x-2x+(12-2x))=(12-2x)(-2x-2x+(12-2x))

=(12-2x)(12-6x)=144-24x-72x+12x=(12-2x)(12-6x)=144-24x-72x+12x22

=12(12-8x+x=12(12-8x+x22))

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-V’=x(12-2x)(-2)+x(-2)(12-2x)+(12-2x)(12-2x)2x)

=(12-2x)(-2x-2x+(12-2x))=(12-2x)(-2x-2x+(12-2x))

=(12-2x)(12-6x)=144-24x-72x+12x=(12-2x)(12-6x)=144-24x-72x+12x22

=12(12-8x+x=12(12-8x+x22) = 12(x-6)(x-2) = 0) = 12(x-6)(x-2) = 0

12” by 12” sheet of 12” by 12” sheet of cardboardcardboard

Find the box with the most Find the box with the most

V’ = 12(12-8x+xV’ = 12(12-8x+x22) = 12(x-6)(x-2) = 0) = 12(x-6)(x-2) = 0

x = 2 or x = 6x = 2 or x = 6

V = (12-2x)(12-2x) x V = (12-2x)(12-2x) x = 144x - 48x= 144x - 48x22 + 4x + 4x33

Find the box with the most Find the box with the most volume.volume.dV/dx =dV/dx =144 – 96 x + 12 x144 – 96 x + 12 x22 = 0 when = 0 when 12(12 - 8x+ x12(12 - 8x+ x22) = 0 ) = 0 (6 – x)(2 – x) = 0(6 – x)(2 – x) = 0

dV/dx = 144 – 96 x + 12 xdV/dx = 144 – 96 x + 12 x22

X = 2 or x = 6X = 2 or x = 6

dd22V/dxV/dx2 =2 =

-96 + 24 x-96 + 24 x

At x = 2 or at x = 6At x = 2 or at x = 6

negative positivenegative positive

dV/dx = 144 – 96 x + 12 xdV/dx = 144 – 96 x + 12 x22

X = 2 or x = 6X = 2 or x = 6

dd22V/dxV/dx2 =2 =

-96 + 24 x-96 + 24 x

At x = 2 or at x = 6At x = 2 or at x = 6

negative positivenegative positive

Local max local minLocal max local min

Steps for solving an Steps for solving an optimization problemoptimization problem

Read the problem drawing a picture as you read Read the problem drawing a picture as you read Label all constants and variables as you read Label all constants and variables as you read If you have two unknowns, write a secondary equation If you have two unknowns, write a secondary equation

Usually the first thing givenUsually the first thing givenFind the variable that you want to optimize and write the primary Find the variable that you want to optimize and write the primary equation equation Eliminate one variable from the primary equation using the Eliminate one variable from the primary equation using the secondary equation secondary equation Determine the domain of the new primary equation Determine the domain of the new primary equation Differentiate the primary equation Differentiate the primary equation

Set the derivative equal to zero Set the derivative equal to zero Solve for the unknown Solve for the unknown

Check the endpoints or run a first or second derivative test Check the endpoints or run a first or second derivative test

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

Semicircle of radius 6. Semicircle of radius 6.

If you have two unknowns, write a If you have two unknowns, write a secondary equation. Usually secondary equation. Usually

the first thing given. the first thing given.

Write the equation of a circle, Write the equation of a circle, centered at the origin of centered at the origin of radius 6.radius 6.

A.A. x + y = 36x + y = 36

B.B. xx22 + y + y22 = 6 = 6

C.C. xx22 + y + y22 = 36 = 36

D.D. y = y = 26 x

We We identify theidentify the primary primary equationequation by the key word by the key word maximizes or minimizesmaximizes or minimizes

Find the value of x that Find the value of x that maximizes the blue area.maximizes the blue area.

Find the rectangle with the Find the rectangle with the largest area largest area

Find the value of x that Find the value of x that maximizes the blue area.maximizes the blue area.

Which of the following Which of the following is the primary is the primary equation?equation?A.A. A = x yA = x y

B.B. A = 2 x yA = 2 x y

C.C. A = ½ x yA = ½ x y

D.D. A = 4 x yA = 4 x y

Eliminate one variable from Eliminate one variable from the primary equation using the primary equation using the secondary equation the secondary equation

A(x) = 2xy = 2x(6A(x) = 2xy = 2x(622 - x - x22 ) )½½

AA22 = 4x = 4x22(36 - x(36 - x22) = 144x) = 144x22 - 4x - 4x44

Differentiate Differentiate AA22 = 144x = 144x2 2 - 4x- 4x44

implicitly.implicitly.

A.A. A’ = 288x - 16xA’ = 288x - 16x33

B.B. 2AA’ = 144x - 8x2AA’ = 144x - 8x

C.C. A’ = 144x – 16xA’ = 144x – 16x

D.D. 2AA' = 288x - 16x2AA' = 288x - 16x33

AA' = 144x - 8xAA' = 144x - 8x33 = 0 = 0Solve for xSolve for x

A.A. x= 0, 3 root(2), - 3 root(2)x= 0, 3 root(2), - 3 root(2)

B.B. x = 6 root(2), - 6 root(2)x = 6 root(2), - 6 root(2)

C.C. x = 0, 3, -3x = 0, 3, -3

D.D. x = 3/root(2), - 3/root(2)x = 3/root(2), - 3/root(2)

Check the endpoints or run Check the endpoints or run a first or second derivative a first or second derivative testtest

AA' = 18x - xAA' = 18x - x3 3 = x(18 – x = x(18 – x22))

A' = 0 when x = 3 root(2) or x = 0A' = 0 when x = 3 root(2) or x = 0

AA’(3)= 54 - 27 > 0AA’(3)= 54 - 27 > 0

AA’(6) = 108AA’(6) = 108 - 6- 63 3 < 0 < 0

AA’(3)= 54 - 27 > 0AA’(3)= 54 - 27 > 0 AA’(6) = 108 AA’(6) = 108 - 6- 63 3 < 0 < 0

A.A. There is a local max at x = 3 There is a local max at x = 3 root(2)root(2)

B.B. Neither a max nor min at 3 Neither a max nor min at 3 root(2)root(2)

C.C. There is a local min at x = 3 There is a local min at x = 3 root(2)root(2)

D.D. Inflection point at x = 3 root(2)Inflection point at x = 3 root(2)

Steps for solving an Steps for solving an optimization problemoptimization problemRead the problem drawing a picture as you read Read the problem drawing a picture as you read Label all constants and variables as you read Label all constants and variables as you read If you have two unknowns, write a secondary equation If you have two unknowns, write a secondary equation

Usually the first thing givenUsually the first thing givenFind the variable that you want to optimize and write the primary Find the variable that you want to optimize and write the primary equation equation Eliminate one variable from the primary equation using the Eliminate one variable from the primary equation using the secondary equation secondary equation Determine the domain of the new primary equation Determine the domain of the new primary equation Differentiate the primary equation Differentiate the primary equation

Set the derivative equal to zero Set the derivative equal to zero Solve for the unknown Solve for the unknown

Check the endpoints or run a first or second derivative test Check the endpoints or run a first or second derivative test

Build a rain gutter Build a rain gutter with the dimensions with the dimensions shown.shown.Base Area = h(b+1)Base Area = h(b+1)

BA = sin(BA = sin()[cos()[cos()+1])+1]

V=V=

BA = sin(BA = sin()[cos()[cos()+1])+1]V=V=

A.A. 20 sin(20 sin()[cos()[cos()+1])+1]

B.B. 20 sin(20 sin())22[cos([cos()+1])+1]22

C.C. sin(sin()[cos()[cos()+1])+1]22

D.D. sin(sin())22[cos([cos()+1])+1]

Find Find that maximizes that maximizes the volumethe volume

V = 20 BAV = 20 BA

V = 20 sin(V = 20 sin()[cos()[cos() + 1 ]) + 1 ]

V’ =V’ =

V=20 sin(V=20 sin()[cos()[cos()+1])+1]dV/ddV/d = =

A.A. 20 sin(20 sin()[cos()[cos()+1])+1]

B.B. 20 cos(20 cos() sin() sin() )

C.C. 20[sin(20[sin()(-sin()(-sin())+(cos())+(cos())+1)cos(+1)cos()])]

D.D. - 20 cos(- 20 cos() sin() sin() )

Find Find that maximizes that maximizes the volumethe volume

V’ = 20sin(V’ = 20sin()[-sin()[-sin()]+[cos()]+[cos() + 1]20 ) + 1]20 cos(cos())

= 20 cos= 20 cos22(() - 20 sin) - 20 sin22(() + 20 cos() + 20 cos())

= 20[2 cos(= 20[2 cos() - 1][cos() - 1][cos() + 1] = 0) + 1] = 0

20[2 cos(20[2 cos() - 1][cos() - 1][cos() + 1] = ) + 1] = 00Solve for Solve for onon..

A.A. /3 or/3 orB.B. /2 or/2 orC.C. /6 or/6 or

D.D. /3 or/3 or

Find Find that maximizes that maximizes the volume of the the volume of the guttergutter

V’ = 20[2 cos(V’ = 20[2 cos() - 1][cos() - 1][cos() + 1] = 0) + 1] = 0

2 cos(2 cos() = 1 or cos() = 1 or cos() = -1) = -1

oror

Find Find that maximizes that maximizes the volume of the the volume of the guttergutter

V’ = 20 cosV’ = 20 cos22(() - 20 sin) - 20 sin22(() + 20 cos() + 20 cos())

V’’ = -40 cos(V’’ = -40 cos()sin()sin() – 40 sin() – 40 sin()cos()cos() -20 ) -20 sin(sin())

V’’(V’’() = -40(½) – 40 (½) - 20) = -40(½) – 40 (½) - 20

a local maximum at x = a local maximum at x = V”(V”() = 0 -> Test fails) = 0 -> Test fails

3

23

2

3

2

Local max at Local max at = = /3/3

V’ = 20 cosV’ = 20 cos22(() - 20 sin) - 20 sin22(() + 20 cos() + 20 cos())

V’(V’(/2) = -20 V’(3/2) = -20 V’(3/2) = -20/2) = -20

Second derivative test failedSecond derivative test failed

First derivative test says decreasing on First derivative test says decreasing on [[/3, /3, ]]

GSU builds 400 GSU builds 400 meter track.meter track.

400 = 400 =

2x + 2x + d d

Soccer requires Soccer requires a maximuma maximumof green areaof green area

A =A =

xd, but d = because 400 = 2x xd, but d = because 400 = 2x + + d d

So A = So A =

400 2x

2400 2x x

Soccer requires Soccer requires a maximuma maximumgreen rectanglegreen rectangle

So A = So A =

and A’ = when x = 100 metersand A’ = when x = 100 meters

A” =A” =

2400 2x x

400 40

x

Soccer requires Soccer requires a maximuma maximumgreen areagreen area

400 = 2x + 400 = 2x + d and when x = 100 meters d and when x = 100 meters

200 = 200 = d or d = 200 / d or d = 200 /