APPLIED PROBLEMS APPLIED PROBLEMS IN MAXIMA AND IN MAXIMA AND

MINIMAMINIMA

MAXIMUM/MINIMUM PROBLEMS MAXIMUM/MINIMUM PROBLEMS

The following problems are maximum/minimum The following problems are maximum/minimum optimization problems. They illustrate one of the most optimization problems. They illustrate one of the most important applications of the first derivative. Many important applications of the first derivative. Many students find these problems intimidating because they students find these problems intimidating because they are "word" problems, and because there does not are "word" problems, and because there does not appear to be a pattern to these problems. However, if appear to be a pattern to these problems. However, if you are patient you can minimize your anxiety and you are patient you can minimize your anxiety and maximize your success with these problems by following maximize your success with these problems by following these guidelines : these guidelines :

GUIDELINES FOR SOLVING MAX./MIN. PROBLEMS GUIDELINES FOR SOLVING MAX./MIN. PROBLEMS 1. Read each problem slowly and carefully. Read the 1. Read each problem slowly and carefully. Read the

problem at least three times before trying to solve it. problem at least three times before trying to solve it. Sometimes words can be ambiguous. It is imperative to Sometimes words can be ambiguous. It is imperative to know exactly what the problem is asking. If you misread know exactly what the problem is asking. If you misread the problem or hurry through it, you have NO chance of the problem or hurry through it, you have NO chance of solving it correctly.solving it correctly.

2. If appropriate, draw a sketch or diagram of the problem 2. If appropriate, draw a sketch or diagram of the problem to be solved. Pictures are a great help in organizing and to be solved. Pictures are a great help in organizing and sorting out your thoughts.sorting out your thoughts.

MAXIMUM/MINIMUM PROBLEMS MAXIMUM/MINIMUM PROBLEMS

3. Define variables to be used and carefully label your 3. Define variables to be used and carefully label your picture or diagram with these variables. This step is very picture or diagram with these variables. This step is very important because it leads directly or indirectly to the important because it leads directly or indirectly to the creation of mathematical equations. creation of mathematical equations.

4. Write down all equations which are related to your 4. Write down all equations which are related to your problem or diagram. Clearly denote that equation which problem or diagram. Clearly denote that equation which you are asked to maximize or minimize. Experience will you are asked to maximize or minimize. Experience will show you that MOST optimization problems will begin show you that MOST optimization problems will begin with two equations. One equation is a "constraint" with two equations. One equation is a "constraint" equation and the other is the "optimization" equation. equation and the other is the "optimization" equation. The "constraint" equation is used to solve for one of the The "constraint" equation is used to solve for one of the variables. This is then substituted into the "optimization" variables. This is then substituted into the "optimization" equation before differentiation occurs. Some problems equation before differentiation occurs. Some problems may have NO constraint equation. Some problems may may have NO constraint equation. Some problems may have two or more constraint equations. have two or more constraint equations.

5. Before differentiating, make sure that the optimization 5. Before differentiating, make sure that the optimization equation is a function of only one variable. Then equation is a function of only one variable. Then differentiate using the well-known rules of differentiation. differentiate using the well-known rules of differentiation.

ExampleExample

PROBLEM 1 :PROBLEM 1 : Find two nonnegative Find two nonnegative numbers whose sum is 9 and so that the numbers whose sum is 9 and so that the product of one number and the square product of one number and the square of the other number is a maximum.of the other number is a maximum.

SOLUTION 1 :SOLUTION 1 : Let variables Let variables xx and and yy represent two nonnegative numbers. represent two nonnegative numbers. The sum of the two numbers is given to The sum of the two numbers is given to bebe

9 = x + y or y = 9 - x9 = x + y or y = 9 - x

We wish to MAXIMIZE the PRODUCT We wish to MAXIMIZE the PRODUCT

PP = = x yx y22 . .

However, before we differentiate the However, before we differentiate the right-hand side, we will write it as a right-hand side, we will write it as a function of function of xx only. Substitute for only. Substitute for yy getting getting

PP = = x yx y22

= = xx ( 9- ( 9-xx))22

Now differentiate this equation using Now differentiate this equation using the product rule and chain rule, getting the product rule and chain rule, getting

P P ' = ' = xx (2) ( 9- (2) ( 9-xx)(-1) + ( 9-)(-1) + ( 9-xx))22(1) = 0 (1) = 0

( 9-( 9-xx) [ -2) [ -2xx + ( 9- + ( 9-xx) ] =0) ] =0( 9-( 9-xx) [ 9-3) [ 9-3xx ] =0 ] =03( 9-3( 9-xx)( 3-)( 3-xx )=0 )=0 9-9-x x =0, 3-=0, 3-xx =0 =0thereforethereforexx=9 or =9 or xx=3 =3

If x=9, y = 9 – x = 9 – 9 = 0, which does If x=9, y = 9 – x = 9 – 9 = 0, which does not satisfy the condition.not satisfy the condition.

It means the two non-negative numbers are It means the two non-negative numbers are 3 and 9 – 3 or 6. 3 and 9 – 3 or 6.

PROBLEM 2 :PROBLEM 2 : Build a rectangular pen Build a rectangular pen with three parallel partitions using with three parallel partitions using 500 feet of fencing. What dimensions 500 feet of fencing. What dimensions will maximize the total area of the will maximize the total area of the pen ? pen ?

SOLUTION 2 :SOLUTION 2 : Let variable Let variable xx be the be the width of the pen and variable width of the pen and variable yy the the length of the pen. length of the pen.

y

y

x x x x x

Illustration

The total amount of fencing is given to be The total amount of fencing is given to be

500 = 5 (width) + 2 (length) = 5500 = 5 (width) + 2 (length) = 5xx + 2 + 2yy , ,

so that so that

22yy = 500 - 5 = 500 - 5xx

or or

yy = 250 - (5/2) = 250 - (5/2)xx . .

We wish to MAXIMIZE the total AREA of the We wish to MAXIMIZE the total AREA of the pen pen

AA = (width) (length) = = (width) (length) = x yx y . .

However, before we differentiate the right-However, before we differentiate the right-hand side, we will write it as a function of hand side, we will write it as a function of xx only. Substitute for only. Substitute for yy getting getting

AA = = x yx y

= = xx ( 250 - ( 250 - 55//22xx) )

= 250= 250xx - ( - (55//22))xx22 . . Now differentiate this equation, getting Now differentiate this equation, getting

A A ' = 250 - (' = 250 - (55//22) 2) 2xx =0 =0250 - 5250 - 5xx =0 =05 (50 - 5 (50 - xx ) =0 ) =0Solving for x, x = 50 ft. Therefor y= (250-Solving for x, x = 50 ft. Therefor y= (250-

55//22x) = 125 ftx) = 125 ft

PROBLEM 3 :PROBLEM 3 : An open rectangular An open rectangular box with square base is to be made box with square base is to be made from 48 ftfrom 48 ft22 of material. What of material. What dimensions will result in a box with dimensions will result in a box with the largest possible volume ? the largest possible volume ?

SOLUTION 3 :SOLUTION 3 : Let variable Let variable xx be the be the length of one edge of the square length of one edge of the square base and variable base and variable yy the height of the the height of the box. box.

x

x

y

Illustration

The total surface area of the box is The total surface area of the box is given to be given to be

48 = (area of base) + 4 (area of one 48 = (area of base) + 4 (area of one side) = side) = x x 22 + 4 ( + 4 (xyxy) , ) ,

so that so that

44xyxy = 48 – = 48 – x x 22

or or

4x

x48y

2

We wish to MAXIMIZE the total VOLUME of the box We wish to MAXIMIZE the total VOLUME of the box VV = (length) (width) (height) = ( = (length) (width) (height) = (xx) () (xx) () (yy) = ) = x x 22yy . . However, before we differentiate the right-hand However, before we differentiate the right-hand

side, we will write it as a function of side, we will write it as a function of xx only. only. Substitute for Substitute for yy getting getting

VV = = x x 22yy

or V = 12x - (1/4)x or V = 12x - (1/4)x 3 3

x

4

1

x

12xV 2

Now differentiate this equation, getting Now differentiate this equation, getting

V V ’= 12 - (’= 12 - (11//44)3)3x x 22 = 0 = 0

12 - (12 - (33//44))x x 22 = 0 = 0

((33//44)(16 – )(16 – x x 22 ) = 0 ) = 0

((33//44)(4 - )(4 - xx)(4 + )(4 + xx) = 0 ) = 0 Solving for x, x Solving for x, x = 4 ft or = 4 ft or x x = -4 ft = -4 ft

(disregarded)(disregarded)

Therefore y = (Therefore y = (1212//44 - - 11//44[4]) = 3 – 1 = 2 [4]) = 3 – 1 = 2 ft and then ft and then

VV = 32 ft = 32 ft33

PROBLEM 4 :PROBLEM 4 : A container in the A container in the shape of a right circular cylinder with shape of a right circular cylinder with no top has surface area 3no top has surface area 3ΠΠftft22. What . What height h and base radius r will height h and base radius r will maximize the volume of the maximize the volume of the cylinder ? cylinder ?

SOLUTION 4 :SOLUTION 4 : Let variable Let variable rr be the be the radius of the circular base and radius of the circular base and variable variable hh the height of the cylinder. the height of the cylinder.

r

h

Illustration

The total surface area of the cylinder is The total surface area of the cylinder is given to be given to be

SA = (area of base) + (area of the SA = (area of base) + (area of the curved side) = 3curved side) = 3ΠΠ

ΠΠrr22 + 2 + 2ΠΠrhrh = 3 = 3ΠΠ

or h = or h = (3(3ΠΠ – – ΠΠr2)r2)//22ΠΠr r = = 33//2r2r - - rr//22

We wish to MAXIMIZE the total We wish to MAXIMIZE the total VOLUME of the cylinder VOLUME of the cylinder

VV = (area of base) (height) = (area of base) (height)

However, before we differentiate the However, before we differentiate the right-hand side, we will write it as a right-hand side, we will write it as a function of function of rr only. Substitute for only. Substitute for hh gettinggetting

V=V=ΠΠrr22((33//2r2r - - rr//22) =) =33ΠΠrr//22 - - ΠΠr3r3//22

Now differentiate this equation, getting Now differentiate this equation, getting

V ‘ = V ‘ = 33ΠΠ//22 - - 33ΠΠr2r2//22 = 0 = 0

33ΠΠ//22 (1 - r (1 - r22) = 0) = 0

33ΠΠ//22 (1 + r)(1 - r) = 0 (1 + r)(1 - r) = 0

Solving for r:Solving for r:

r = 1 ft or r = -1 (disregarded) r = 1 ft or r = -1 (disregarded)

and and = = 33//2(1)2(1) - - 11//2 2 = = 33//22 - - 11//2 2 = = 22//22 =1 ft =1 ft

PROBLEM 5 :PROBLEM 5 : A sheet of cardboard 3 A sheet of cardboard 3 ft. by 4 ft. will be made into a box by ft. by 4 ft. will be made into a box by cutting equal-sized squares from cutting equal-sized squares from each corner and folding up the four each corner and folding up the four edges. What will be the dimensions edges. What will be the dimensions of the box with largest volume ? of the box with largest volume ?

SOLUTION 5 :SOLUTION 5 : Let variable Let variable xx be the be the length of one edge of the square cut length of one edge of the square cut from each corner of the sheet of from each corner of the sheet of cardboard. cardboard.

4

3

Illustration

x

x x

x

x

x

x

x

4-2x

3-2x

x

After removing the corners and folding After removing the corners and folding up the flaps, we have an ordinary up the flaps, we have an ordinary rectangular box. rectangular box.

We wish to MAXIMIZE the total VOLUME of the We wish to MAXIMIZE the total VOLUME of the box box

VV = (length) (width) (height) = (4-2 = (length) (width) (height) = (4-2xx) (3-2) (3-2xx) () (xx) ) Now differentiate this equation using the triple Now differentiate this equation using the triple

product rule, getting product rule, getting V V ' = (4-2' = (4-2xx)(3-2)(3-2xx)(1)+(4-2)(1)+(4-2xx)(x)(-2)+(3-2)(x)(-2)+(3-2xx)()(xx)(-)(-

2)=02)=0Solving for x:Solving for x:12-14x+4x12-14x+4x22-8x-8x +4x+4x22-6x+4x-6x+4x22=0=012x12x22-28x-28x +12=0+12=04(3x4(3x22-7x-7x +3)=0 +3)=0 Using the quadratic equation:Using the quadratic equation:

ft 0.57or ed)(disregard 1.776

137x

Exercises. Solve each of the Exercises. Solve each of the following:following:

PROBLEM 6 :PROBLEM 6 : Consider all triangles formed Consider all triangles formed by lines passing through the point (by lines passing through the point (88//99,3) and ,3) and both the x- and y-axes. Find the dimensions both the x- and y-axes. Find the dimensions of the triangle with the shortest hypotenuse. of the triangle with the shortest hypotenuse.

PROBLEM 7 :PROBLEM 7 : Find the point (x, y) on the Find the point (x, y) on the graph of nearest the point (4, 0). graph of nearest the point (4, 0).

PROBLEM 8 :PROBLEM 8 : A cylindrical can is to hold A cylindrical can is to hold 2020ΠΠmm33 The material for the top and bottom The material for the top and bottom costs $10/mcosts $10/m22 and material for the side costs and material for the side costs $8/m$8/m22 Find the radius r and height h of the Find the radius r and height h of the most economical can. most economical can.

xy

PROBLEM 9 :PROBLEM 9 : You are standing at the edge of a slow- You are standing at the edge of a slow-moving river which is one mile wide and wish to moving river which is one mile wide and wish to return to your campground on the opposite side of return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on You must first swim across the river to any point on the opposite bank. From there walk to the the opposite bank. From there walk to the campground, which is one mile from the point campground, which is one mile from the point directly across the river from where you start your directly across the river from where you start your swim. What route will take the least amount of time ? swim. What route will take the least amount of time ?

PROBLEM 10 :PROBLEM 10 : Construct a window in the shape of a Construct a window in the shape of a semi-circle over a rectangle. If the distance around semi-circle over a rectangle. If the distance around the outside of the window is 12 feet, what the outside of the window is 12 feet, what dimensions will result in the rectangle having largest dimensions will result in the rectangle having largest possible area ? possible area ?

PROBLEM 11 :PROBLEM 11 : There are 50 apple trees in an There are 50 apple trees in an orchard. Each tree produces 800 apples. For orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, each additional tree planted in the orchard, the output per tree drops by 10 apples. How the output per tree drops by 10 apples. How many trees should be added to the existing many trees should be added to the existing orchard in order to maximize the total orchard in order to maximize the total output of trees ? output of trees ?

PROBLEM 12 :PROBLEM 12 : Find the dimensions of the Find the dimensions of the rectangle of largest area which can be rectangle of largest area which can be inscribed in the closed region bounded by inscribed in the closed region bounded by the x-axis, y-axis, and graph of the x-axis, y-axis, and graph of y y = 8 - = 8 - x x 33

PROBLEM 13 :PROBLEM 13 : Consider a rectangle of Consider a rectangle of perimeter 12 inches. Form a cylinder by perimeter 12 inches. Form a cylinder by revolving this rectangle about one of its edges. revolving this rectangle about one of its edges. What dimensions of the rectangle will result in What dimensions of the rectangle will result in a cylinder of maximum volume ? a cylinder of maximum volume ?

PROBLEM 14 :PROBLEM 14 : A movie screen on a wall is 20 A movie screen on a wall is 20 feet high and 10 feet above the floor. At what feet high and 10 feet above the floor. At what distance x from the front of the room should distance x from the front of the room should you position yourself so that the viewing angle you position yourself so that the viewing angle of the movie screen is as large as possible ? of the movie screen is as large as possible ?

PROBLEM 15 :PROBLEM 15 : Find the dimensions (radius r Find the dimensions (radius r and height h) of the cone of maximum volume and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2. which can be inscribed in a sphere of radius 2.