4.4 Optimization I. - Tidewater Community College

Transcript 4.4 Optimization I. - Tidewater Community College

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 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.

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.

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.

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 Two variables that’s a problem 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 But that’s where the constraint equation comes into play: Now substitute for l in: A= l w A(w)= (25-w)w By solving it for l or w we can eliminate one of the unknowns.

50 -2w = 2l 25 - w = l Now Area is only in terms of one variable w and we can differentiate.

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?

12.5

A rectangle that is a square.

12.5

12.5 ft.

The perimeter is 12.5 (4)= 50 ft.

of fencing.

Area is 156.25 sq. ft. or ft 2 How do we know this is the maximum area?

y 12.5 ft.

A quick way is to look at the graph of our function A(w) = 25w - w 2 50 40 30 20 10 90 80 70 60 x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 At x = 12.5 we reach the peak of the curve or the vertex.

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

Width W Length L Area A Max is 156.25

10 13 12.25

2L +2(10)=50 2L + 20= 50 2L = 30 L = 15 2L +2(13)=50 2L + 26= 50 2L = 24 L = 12 2L +2(12.25)=50 2L + 24.5= 50 2L = 25.5

L = 12.75

Area 10(15) = 150 13(12)=156 12.25(12.75)= 156.1875

Guidelines for Solving Optimization Problems 1. Assign a letter to each variable mentioned in the problem. If appropriate, draw and label a figure.

2. Find an expression for the quantity to be optimized.

3. 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 4. Optimize the function f over its domain using the methods of Section 4.4

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?

2 



l V

 

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.

In this problem the formulas are given in the hint.

Can you write the equation to be maximized?

2 



l V

 

In this problem the formulas are given in the hint.

Can you write the equation to be maximized?

 

r l

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

How many variables are in the Volume formula?

2 



l V

 

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?

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

2 



l V

 

What is the constraint equation?

2 



= 108 Which is the easiest variable to solve for r or l ?

 108  2 

The equation to maximize.

V V V

  



2 2

 108  2 

 (

)  108 

2  2  2

3 The constraint equation

 108  2 

Finding the derivative of V( r)

(

)  108 

2  2  2

0  (

 )  6 

216 

 36   

 6  2

2 6 

r r

  0 0 , 36 , 36   



r r

 0 ,

 36   0 The critical points are 0 and 36  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.

 (

)  216 

 6  2

V V

 (

)  216   12  2

  36   216   12  2 36   216   432    216 

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

2 



l V

 

l l

We have found the dimension of the radius to be  108  2  36  The volume is

  108   36   2 72 36   36 36 3   46 , 656   14851 .

06605

3 36  Is the length of the package.

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.

16 x x 10

10 16 x x x 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.

We are constrained by the size of the paper we are using.

The volume of a rectangular solid is V= LWH (length, width, height) V(x) = (16-2x)(10-2x)x

(

)  ( 16  2

)( 10  2

)

x v

(

)

(

)   ( 160  160   32

  20 4



 4

2 )

x v

(

)  160

 52

2  4

3 To maximize the volume we take v’(x).

(

)  160

 52

2  4

 (

)  160  104

 12

2 0  12

2  104

 160

divide

_ 4 0  3

2  26

 40

Now we solve for the value x has to be v ( x )  160 x  52 x 2  4 x 3 0 0 3 x   3 x  3 x 2   26 x 20  x   40 2   20 , x  2 x  20 3  6 2 3 _ or _ x  2 In order to determine which of these numbers gives a max or a min we can use the second derivative test.

v ( x )  160 x  52 x 2  4 x 3 v  ( x ) v  ( x ) v  ( 2 )    160  104 x  104  104  12 x 2   24 x 24     104 Since _ v  ( 2 )  0 , we _ have  48   56 _ a _ max v  ( 20 3 ) Since _  v  104  ( 20 3 )  24     0 , 20 3 we      _ have 104 _ a  160 _ min  56

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

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

X 16-2x 10-2x Volume=x(16-2x)(10-2x) 1.5 16-3=13 10-3=7 136.5

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

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

Minimizing Cost Example 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 36in 3 and is constructed using the least amount of metal.

To get some information on Volume see the next slide

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

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.

V   r 2 in that h h r Click for sound

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 36in amount of metal.

3 and is constructed using the least h What is the constraint?

The volume.

The material which makes up the can.

3 and is constructed using the least h r What is the constraint?

If you chose The volume.

Right, 36in 3 to be precise.

The material which makes up the can.

No, this is what we want to minimize.

3 and is constructed using the least h r Write a formula for the volume and, Write a formula for the material which makes up the can.

Wait to click till you do.

3 and is constructed using the least h r Write a formula for the volume and, V   r 2 h Write a formula for the material which makes up the can.

 r Which totals = 2  r 2  r  2  rh 2  rh

Explanation of formulas: V   r 2  r 2 h Is the area of the circle.

Multiply that by the height to get Volume.

h r  r  r 2  rh The can has a top and bottom whose area is a circle thus 2  r 2 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 2  r 2  rh Added together = 2  r 2  2  rh

The Volume constraint is written:

36



Solve



r

h

for 36



r

2 

h

The equation to minimize is: derivative taken.

f ( r , h )  2  r 2  2  rh which has two variables r and h and as such cannot yet have its 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 ).

f ( r )  2  r 2 f ( r )  2  r 2  2  r 36  r 2  72 r  2  r 2  72 r  1

To minimize we find f’(r ) and set equal to 0.

f ( r )  2  r 2  72 r  2  r 2  72 r  1 f  72 r r r 3 3 (  r 0  0  4  r 4  r 3   72 r 2 72   3 )   18  4  r 4  r 3 72 4 18    72 r  2  4  r  72 r 2 f  ( r )  4  r  72 r  2 To check that this value does create a minimum, lets do the 2nd derivative test.

f  ( r )  f    3 18  4     4 144 r    3  4   144 r 3   3 144 18    3  4   144 18   4   8   12  Which is positive and thus creates a minimum.

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

h



36



r

2      3

36 18

    2     

36 18

    2 3  

36

  2 3  2 3  3 1

36

  2 3   3 1

2



18

  2 3 

2

   1 3 1  3 

2

18



This is what we have found : The can has a capacity of 36 in The radius and height which will give the minimum amount of material used is r  3 18 

h

3 

2

18

 First, let’s check that the volume is 36 in 3 .

V     3 18   2  2 3 18    2  18   36

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

3 18   1 .

789 ...

r 1.5

3 18  2 h  36  r 2 36  ( 2 .

25 ) 2 3 18  h  36  4  9  V   r 2 h V   2 .

25 2 .

36 25   36 36 V   4   9   36 2  r 2  2  rh 2   18  2  4 .

5  2  .

25   48  2     62 .

2 137 36 .

25  2 3 2 2    r  2 1 3 2    18    2 1   6   rh   2 3  60 .

355 2  4  2    9   8   36  61 .

132

Good Luck on the problems!

4.4 Optimization I. - Tidewater Community College

Transcript 4.4 Optimization I. - Tidewater Community College

4.5 Optimization II

right circular cylinder

36

Solve

r

h

for 36

r

h

h

h

36

r

36 18

36 18

36

36

2

18

2

2

18

h

2

18

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