Applied max and minApplied max and min

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 = 144x - 48x2 +4x3

V = 144x - 48x2 +4x3

V’ =A. 144 - 96 x + 12B. 122 -96 x + 12 x2 C. 144 – 96 x + 12 x2

V = 144x - 48x2 +4x3

Find the box with the most volume.dV/dx =

144 – 96 x + 12 x2 = 0 when 12(12 - 8x+ x2) = 0

0 = 144 – 96 x + 12 x2

When 12 - 8x+ x2 = 0A. x = 0 or x = 6B. x = 2 or x = 4C. x = 2 or x = 6

dV/dx = 144 – 96 x + 12 x2

12(6 – x)(2 – x) = 0X = 2 or x = 6

d2V/dx2 =

-96 + 24 xAt x = 2 and at x = 6negative positive

Steps for solving an Steps for solving an optimization problemoptimization problem

Write a secondary equation and solve Write a secondary equation and solve for yfor yWrite the primary equation and replace Write the primary equation and replace y y 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 Check the endpoints or run a first or second derivative test 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 construct a rectangle.construct a rectangle.

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.

What is the area of her What is the area of her garden?garden?

A = L * WA = L * WA=5 * 990A=5 * 990A = 4950 sq. ft.A = 4950 sq. ft.

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 given2x + y = 1000 is the secondary2x + y = 1000 is the secondarySolve for ySolve for yy = 1000 – 2xy = 1000 – 2x

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 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

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 = 0A’’ = -4 concave downA’’ = -4 concave downA’’(250) = -4A’’(250) = -4Relative max at x = 250Relative max at x = 250A = 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 volumeA. 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.

A box with a square A box with a square base and no top holds base and no top holds 100 in100 in33.. 100 =100 = xx22*y *y Minimize the construction costMinimize the construction cost Cost is based upon ___ areas.Cost is based upon ___ areas. A = A = 4xy+4xy+ xx2 2 = 4x[100/ x= 4x[100/ x22] + x] + x22

A = 4x[100/ xA = 4x[100/ x22] + x] + x22

= 400/x + x= 400/x + x22 A’(x) A’(x) = =

A(x) = 400 xA(x) = 400 x-1-1 + x + x22 A’(x) = A’(x) =A.A. -400 x-400 x00 + 2x + 2xB.B. 400 x400 x-2-2 + 2x + 2xC.C. -400 x-400 x-2-2 + 2x + 2xD.D. -400 x-400 x-2-2 - 2x - 2x

A’(x) = -400 xA’(x) = -400 x-2-2 + 2x=0 + 2x=0 when x = when x = A.A. x = 200x = 200B.B. x =x =C.C. x =x =D.D. x = x =

2003 2004 200

GSU builds newGSU builds new400 meter track.400 meter track.400 = 400 =

Circumference of a whole Circumference of a whole circle is circle is d so for a 400 d so for a 400 meter trackmeter track

A.A. 400 = x * 400 = x * ddB.B. 400 = x + 400 = x + ddC.C. 400 = 2x + 400 = 2x + ddD.D. 400 = 2x + 2400 = 2x + 2dd

Solve for dSolve for d400 = 2x + 400 = 2x + ddA. d =

.B. d =

.C. d =

400 2x

400 2x

2400 x

Soccer requires a Soccer requires a maximummaximumgreen area A = L * Wgreen area A = L * WA.A. A = 2x * dA = 2x * dB.B. A = x * dA = x * dC.C. A = x + dA = x + dD.D. A = 2x + 2dA = 2x + 2d

A = x * d but d = A = x * d but d =

A.A. A = A = B.B. A =A =C.C. A =A =

400 2 (400 2 ) /x x

2(400 2 ) /x x (400 2 ) /x x (400 2) /x

A = A’ A = A’ = = A.A. A’ = A’ = B.B. A’ = A’ = C.C. A’ = A’ =

2(400 2 ) /x x

(400 4 ) /x x (400 4 ) /x (400 4 ) /x

A’ = = 0 A’ = = 0 when when A.A. x = 100 metersx = 100 metersB.B. x = 200 metersx = 200 metersC.C. x = 400 metersx = 400 meters

(400 4 ) /x

Soccer requires Soccer requires a maximuma maximumgreen rectanglegreen rectangle

So A = So A =

and A’ = when x = 100 metersand A’ = when x = 100 metersA” =A” =

2400 2x x

400 4 0x

A’ = A’’ A’ = A’’ (100)= (100)=

A.A. A’’(100) = - 4/A’’(100) = - 4/B.B. A’’(100) = 0A’’(100) = 0C.C. A’’(100) = - 400/A’’(100) = - 400/

400 4 (400 4 ) /x x

Since A’’(100) < 0 A Since A’’(100) < 0 A has a relativehas a relativeA.A. minimum at x = 100 metersminimum at x = 100 metersB.B. maximum at x = 100 metersmaximum at x = 100 metersC.C. Neither! The second derivative Neither! The second derivative

test fails at x = 100 meterstest fails at x = 100 meters

Soccer requires Soccer requires a maximuma maximumgreen areagreen area

AA is a maximum when is a maximum when 400 = 2x + 400 = 2x + d and when x = 100 d and when x = 100 metersmeters200 = 200 = d or d = 200 / d or d = 200 /

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 = 36B.B. xx22 + y + y22 = 6 = 6C.C. xx22 + y + y22 = 36 = 36D.D. y = y = 26 x

This is the secondary This is the secondary equation. xequation. x22 + y + y22 = = 3636

Solving for ySolving for y22 we get we getyy22 = 36 - x = 36 - x22

or y = or y = 236 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 yB.B. A = 2 x yA = 2 x yC.C. A = ½ x yA = ½ x yD.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 - 8xC.C. A’ = 144x – 16xA’ = 144x – 16xD.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, -3D.D. x = 3/root(2), - 3/root(2)x = 3/root(2), - 3/root(2)

Run a first derivative Run a first 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 = 0AA’(3)= 54 - 27 > 0AA’(3)= 54 - 27 > 0AA’(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 = There is a local max at x = B.B. Neither a max nor min at x = Neither a max nor min at x = C.C. There is a local min at x =There is a local min at x =

3 23 2

3 2

The blue areaThe blue areais a max whenis a max when x =x = xx22 + y + y22 = 36 = 36 18 + y18 + y22 = 36 = 36 y =y =

3 2

3 2