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4.5 Optimization II. Dr. Julia Arnold using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text. Suppose you want to make a rectangular garden but you can

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4 5 optimization ii l.jpg

4.5 Optimization II

Dr. Julia Arnold

using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text


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Suppose you want to make a rectangular garden but you can

only afford to buy 50 feet of fencing. What would be the largest possible rectangle that you could have.

This problem is different from the ones in the last section in that you want to optimize one thing (find max or min) but you have been given a constraint (limited amount of fencing in this case).

We need to consider two formulas:

one for the perimeter of the rectangle (which represents the fencing) and one for the area of the rectangle (which represents the largest size garden.


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Suppose you want to make a rectangular garden but you can

only afford to buy 50 feet of fencing. What would be the largest possible rectangle that you could have.

We need to consider two formulas:

one for the perimeter of the rectangle (which represents the fencing) and one for the area of the rectangle (which represents the largest size garden.

This problem is different from the ones in the last section in that you want to optimize one thing (find max or min) but you have been given a constraint (limited amount of fencing in this case).

The formula for the area which we would like to be a maximum is

A = lw

The formula for the perimeter which is the constraint is

50 = 2l + 2w

Since the Area formula is the one for which we seek a maximum, it is the one that we need to find the derivative. But, it has two variables l and w. That’s a problem.


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Two variables

that’s a problem

But that’s where the constraint equation comes into play:

By solving it for l or w we can eliminate one of the unknowns.

50 -2w = 2l

25 - w = l

Suppose you want to make a rectangular garden but you can

only afford to buy 50 feet of fencing. What would be the largest possible rectangle that you could have.

A= l w

The formula for the area which we would like to be a maximum is

A = lw

The formula for the perimeter which is the constraint is

50 = 2l + 2w

Now substitute for l in:

A= l w

A(w)= (25-w)w

Now Area is only in terms of one variable w and we can differentiate.


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The formula to maximize

The constraint equation

A= l w

A(w)= (25-w)w

50 -2w = 2l

25 - w = l

Let’s find A’(w)

A’(w) = (25 - w)1 + w(-1) product rule

Simplify

A’(w) = 25 - w - w = 25 - 2w

Set = 0

0 = 25 - 2w

2w = 25

w = 12.5

To find l go back to the constraint equation and substitute the w you just found.

50 -2w = 2l

25 - w = l

25 - 12.5 = 12.5 = l

What kind of rectangle has l = 12.5 and w = 12.5?


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A rectangle that is a square.

12.5

The perimeter is 12.5 (4)= 50 ft.

of fencing.

Area is 156.25 sq. ft. or ft2

12.5 ft.

12.5

How do we know this is the maximum area?

12.5 ft.

A quick way is to look at the graph of our function

A(w) = 25w - w2

At x = 12.5 we reach the peak of the curve or the vertex.


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Another way to convince ourselves that we have the maximum area is to compute the area of some other rectangles with perimeters of 50.


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Guidelines for Solving Optimization Problems area is to compute the area of some other rectangles with perimeters of 50.

  • Assign a letter to each variable mentioned in the problem. If appropriate, draw and label a figure.

  • Find an expression for the quantity to be optimized.

  • Use the conditions given in the problem (the constraint) to write the quantity to be optimized as a function f of one variable. Note any restrictions to be placed on the domain of f from physical considerations of the problem

  • Optimize the function f over its domain using the methods of Section 4.4


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Example 1: Parcel Post Regulations area is to compute the area of some other rectangles with perimeters of 50.

Postal regulations specify that a parcel sent by parcel post may have a combined length and girth of no more than 108 in. Find the dimensions of the cylindrical package of greatest volume that may be sent through the mail. What is the volume of such a package?

Hint: The length plus the girth is and the volume is

First draw (as best you can) a picture of the cylindrical package.

Next, label the picture: r is the radius of the circular ends and l is the length of the cylinder.

r

In this problem the formulas are given in the hint.

Can you write the equation to be maximized?


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Example 1: Parcel Post Regulations area is to compute the area of some other rectangles with perimeters of 50.

Postal regulations specify that a parcel sent by parcel post may have a combined length and girth of no more than 108 in. Find the dimensions of the cylindrical package of greatest volume that may be sent through the mail. What is the volume of such a package?

Hint: The length plus the girth is and the volume is

In this problem the formulas are given in the hint.

Can you write the equation to be maximized?

If you said: you would be correct as it says,” Find the dimensions of the cylindrical package of greatest volume” and V would stand for volume.

r

How many variables are in the Volume formula?


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Example 1: Parcel Post Regulations area is to compute the area of some other rectangles with perimeters of 50.

Postal regulations specify that a parcel sent by parcel post may have a combined length and girth of no more than 108 in. Find the dimensions of the cylindrical package of greatest volume that may be sent through the mail. What is the volume of such a package?

Hint: The length plus the girth is and the volume is

How many variables are in the Volume formula?

Two r and l.

If there are two what did you learn from the earlier problem about the garden?

r

Two variables is a problem. But the constraint equation can fix it.


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= 108 area is to compute the area of some other rectangles with perimeters of 50.

Example 1: Parcel Post Regulations

Postal regulations specify that a parcel sent by parcel post may have a combined length and girth of no more than 108 in. Find the dimensions of the cylindrical package of greatest volume that may be sent through the mail. What is the volume of such a package?

Hint: The length plus the girth is and the volume is

What is the constraint equation?

Which is the easiest variable to solve for

r or l ?

r


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The equation to maximize. area is to compute the area of some other rectangles with perimeters of 50.

The constraint equation

The critical points are 0 and

Finding the derivative of V( r)

0 creates a minimum because the radius of 0 would mean no cylinder. How do we know that creates a maximum?

Let’s use the 2nd derivative test.


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Since the 2nd derivative is negative that means the function is concave down which means the critical point is creating a maximum.

Example 1: Parcel Post Regulations

Postal regulations specify that a parcel sent by parcel post may have a combined length and girth of no more than 108 in. Find the dimensions of the cylindrical package of greatest volume that may be sent through the mail. What is the volume of such a package?

Hint: The length plus the girth is and the volume is

We have found the dimension of the radius to be

Is the length of the package.

The volume is


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16 is concave down which means the critical point is creating a maximum.

x

x

10

Example 2: By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, the cardboard may be turned into an open box. If the cardboard is 16 inches long and 10 inches wide, find the dimensions of the box that will yield the maximum volume.


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16 is concave down which means the critical point is creating a maximum.

x

x

10

Can you figure out the inside dimensions?

Hint: The long one is 16 – 2x.

The short one is 10 – 2x.

We need this information because they form the dimensions of the box when you cut out the corners and bend up the sides.

x

10-2x

16-2x


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We are constrained by the size of the paper we are using. is concave down which means the critical point is creating a maximum.

The volume of a rectangular solid is

V= LWH (length, width, height)

V(x) = (16-2x)(10-2x)x

To maximize the volume we take v’(x).


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Now we solve for the value x has to be is concave down which means the critical point is creating a maximum.

In order to determine which of these numbers gives a max or a min we can use the second derivative test.


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X 16-2x 10-2x Volume=x(16-2x)(10-2x)

1.5 16-3=13 10-3=7 136.5

144

2 16-4=12 10-4=6 144

2.5 16-5=11 10-5=5 137.5

Thus when x = 2 we get the box with maximum volume.

Just for fun lets compute a few volumes for different x values.

As you can see x=2 produces the largest volume.


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Minimizing Cost Example Volume=x(16-2x)(10-2x)

For its beef stew, the Betty Moore Company uses tin containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in3 and is constructed using the least amount of metal.

To get some information on Volume see the next slide


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h Volume=x(16-2x)(10-2x)

r

A cylinder is what we might think of as a can. While we may have in mathematics slanted cans, the ones in the store are what we call a right circular cylinder in that the sides are perpendicular to the horizontal. The base and top of the can is a circle and thus has a radius r, the distance between the top and bottom is called the height of the can or h. If cut and straightened out this shape would be a rectangle.

Click

for

sound


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h Volume=x(16-2x)(10-2x)

r

For its beef stew, the Betty Moore Company uses tin containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in3 and is constructed using the least amount of metal.

What is the constraint?

The volume.

The material which makes up the can.


Slide23 l.jpg

h Volume=x(16-2x)(10-2x)

r

For its beef stew, the Betty Moore Company uses tin containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in3 and is constructed using the least amount of metal.

What is the constraint?

If you chose

The volume.

Right, 36in3 to be precise.

The material which makes up the can.

No, this is what we want to minimize.


Slide24 l.jpg

h Volume=x(16-2x)(10-2x)

r

For its beef stew, the Betty Moore Company uses tin containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in3 and is constructed using the least amount of metal.

Write a formula for the volume and,

Write a formula for the material which makes up the can.

Wait to click till you do.


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h Volume=x(16-2x)(10-2x)

r

For its beef stew, the Betty Moore Company uses tin containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in3 and is constructed using the least amount of metal.

Write a formula for the volume and,

Write a formula for the material which makes up the can.

Top = bottom = sides =

Which totals =


Slide26 l.jpg

h Volume=x(16-2x)(10-2x)

r

Top = bottom = sides =

Explanation of formulas:

Is the area of the circle.

Multiply that by the height to

get Volume.

The can has a top and bottom whose area is a circle thus

The side of the can rolls out into a rectangle in which the width is h and the length is the same as the Circumference of the Circle which is . The area of a rectangle is LW =

Added together =


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The Volume constraint is written: Volume=x(16-2x)(10-2x)

The equation to minimize is:

which has two variables r and h and as such cannot yet have its

derivative taken.

Next we need to substitute our value for h in the constraint equation into the f(r,h) which would then make it just f(r ).


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To minimize we find f’(r ) and set equal to 0. Volume=x(16-2x)(10-2x)

To check that this value does create a minimum, lets do the 2nd derivative test.

Which is positive and thus creates a minimum.


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For its beef stew, the Betty Moore Company uses tin containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in3 and is constructed using the least amount of metal.

We also need to find the height of the can with this radius.


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This is what we have found : containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in

The can has a capacity of 36 in3

The radius and height which will give the minimum amount of material used is

First, let’s check that the volume is 36 in3.


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r containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in

1.5

2

Now lets check to see if the material used is a minimum amount by choosing values for r around the value we obtained.

36


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Good Luck on the problems! containers that have the form of a right circular cylinder (or a can). Find the radius and the height of a container if it has a capacity of 36in


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