+ All Categories
Home > Documents > Optimization. In optimization problems we are trying to find the maximum or minimum value of a...

Optimization. In optimization problems we are trying to find the maximum or minimum value of a...

Date post: 18-Dec-2015
Category:
Upload: melvin-barrett
View: 217 times
Download: 2 times
Share this document with a friend
Popular Tags:
19
Optimization
Transcript
Page 1: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

Optimization

Page 2: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

In optimization problems we are trying to find the maximum or

minimum value of a variable. The solution is called the optimum

solution.

Page 3: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

Optimization Problem Solving Method

Step 1: If possible, draw a large, clear diagram. Sometimes more than one diagram is necessary

Page 4: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

Optimization Problem Solving Method

Step 2: Construct an equation with the variable to be optimized (maximized or minimized) as the subject of the formula(the y in your calculator) in terms of one convenient variable, x. Find any restrictions there may be on x.

Page 5: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

Optimization Problem Solving Method

Step 3: Find the first derivative and find the value(s) of x when it is zero

Page 6: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

Optimization Problem Solving Method

Step 4: If there is a restricted domain such as axb, the maximum/minimum value of the function may occur either when the derivative is zero or at x=a or at x=b. Show by a sign diagram that you have a maximum or minimum situation.

Page 7: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

A industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The walls, internal and

external, will cost $60 per meter to build. What dimensions would the shed have to minimize the

cost of the walls?

Step 1: If possible, draw a large, clear diagram. Sometimes more than one diagram is necessary.

x m

y m

Page 8: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

Step 2: Construct an equation with the variable to be optimized as the subject of the formula in terms of one convenient variable, x. Find any restrictions there may be on x.

COST

COST=60(Total Length of the Walls)

Page 9: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

Step 2: Construct an equation with the variable to be optimized as the subject of the formula in terms of one convenient variable, x. Find any restrictions there may be on x.

L=6x+4y

C=60(6x+4y)

Page 10: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

Step 2: Construct an equation with the variable to be optimized as the subject of the formula in terms of one convenient variable, x. Find any restrictions there may be on x.

Area=600m2 =3xy

Page 11: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

Step 2: Construct an equation with the variable to be optimized as the subject of the formula in terms of one convenient variable, x. Find any restrictions there may be on x.

Area=600m2=3xy

Solve for y (to get y in terms of x)

y=200 x

Substitute for y in cost formula

C=60(6x+4(200)) x

Page 12: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

Step 2: Construct an equation with the variable to be optimized as the subject of the formula in terms of one convenient variable, x. Find any restrictions there may be on x.

C=60(6x+4(200)) x

YES - x0 and y0

Page 13: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

C=60(6x+4(200)) x

Step 3: Find the first derivative and find the value(s) of x when it is zero

C=360x+48000x-1

C’=360-48000x-2=360-48000 x2

0=360-48000 x2

360=48000 x2

360x2=48000

x2133.3x11.547

Page 14: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

Step 4: If there is a restricted domain such as axb, the max/min value of the function may occur either when the derivative is 0 or at x=a or at x=b. Show by a sign diagram that you have a max or min.

The end points are not going to be where cost is minimized, and where the derivative=0 is a min

Page 15: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

x11.547

Step 4: If there is a restricted domain such as axb, the max/min value of the function may occur either when the derivative is 0 or at x=a or at x=b. Show by a sign diagram that you have a max or min.

+ –

Could see this by putting derivative in your calculator and looking at the table

Page 16: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An industrial shed is to have a total floor space of 600 m2 and is to be divided into 3 rectangular rooms of equal size. The

walls, internal and external, will cost $60 per meter to build. What dimensions would the shed have to minimize the cost of

the walls?x m

y m

What are the dimensions?

3x meters by y m

3x meters by (200/x) m

3(11.547) meters by (200/11.547) m

34.6 meters by 17.3 m

Page 17: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

An open rectangular box has square base and a fixed outer surface area of 108 cm2. What size

must the base be for maximum volume?

ANSWER:6 cm by 6 cm

or36 cm2

Page 18: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

Solve using derivatives: Square corners are cut from a piece of 20 cm by 42 cm cardboard which is then bent into the form of an open box. What size squares should be removed if the volume is

to be maximized?

YOU DO:

Page 19: Optimization. In optimization problems we are trying to find the maximum or minimum value of a variable. The solution is called the optimum solution.

TICKET OUT

A closed box has a square base of side x and height h.

Write down an expression for the volume, V, of the box

Write down an expression for the total surface area, A, of the box

The volume of the box is 1000 cm3

Express h is terms of x

Write down the formula for total surface area in terms of x

Find the derivative (dA/dx)

Calculate the value of x that gives a minimum surface area

Find the surface area for this value of x


Recommended