maximum and minimum

MAXIMUM/MINIMUM PROBLEMSThe following problems are maximum/minimum optimization problems. They illustrate one of the most important applications of the first derivative. Many students find these problems intimidating because they are "word" problems, and because there does not appear to be a pattern to these problems. However, if you are patient you can minimize your anxiety and maximize your success with these problems by following these guidelines :GUIDELINES FOR SOLVING MAX./MIN. PROBLEMS1. Read each problem slowly and carefully. Read the problem at least three times before trying to solve it. Sometimes words can be ambiguous. It is imperative to know exactly what the problem is asking. If you misread the problem or hurry through it, you have NO chance of solving it correctly.2. If appropriate, draw a sketch or diagram of the problem to be solved. Pictures are a great help in organizing and sorting out your thoughts.3. Define variables to be used and carefully label your picture or diagram with these variables. This step is very important because it leads directly or indirectly to the creation of mathematical equations.4. Write down all equations which are related to your problem or diagram. Clearly denote that equation which you are asked to maximize or minimize. Experience will show you that MOST optimization problems will begin with two equations. One equation is a "constraint" equation and the other is the "optimization" equation. The "constraint" equation is used to solve for one of the variables. This is then substituted into the "optimization" equation before differentiation occurs. Some problems may have NO constraint equation. Some problems may have two or more constraint equations.5. Before differentiating, make sure that the optimization equation is a function of only one variable. Then differentiate using the well-known rules of differentiation.6. Verify that your result is a maximum or minimum value using the first or second derivative test for extrema.

The following problems range in difficulty from average to challenging.PROBLEM 1:Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.SOLUTION 1:Let variablesxandyrepresent two nonnegative numbers. The sum of the two numbers is given to be9 =x+y,so thaty= 9 -x.We wish to MAXIMIZE the PRODUCTP=x y2.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygettingP=x y2=x( 9-x)2.Now differentiate this equation using the product rule and chain rule, gettingP' =x(2) ( 9-x)(-1) + (1) ( 9-x)2= ( 9-x) [ -2x+ ( 9-x) ]= ( 9-x) [ 9-3x]= ( 9-x) (3)[ 3-x]= 0forx=9 orx=3 .Note that since bothxandyare nonnegative numbers and their sum is 9, it follows that. See the adjoining sign chart forP' .

Ifx=3 andy=6 ,thenP= 108is the largest possible product.

PROBLEM 2:Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen ?SOLUTION 2:Let variablexbe the width of the pen and variableythe length of the pen.

The total amount of fencing is given to be500 = 5 (width) + 2 (length)= 5x+ 2y,so that2y= 500 - 5xory= 250 - (5/2)x.We wish to MAXIMIZE the total AREA of the penA= (width) (length) =x y.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygettingA=x y=x( 250 - (5/2)x)= 250x- (5/2)x2.Now differentiate this equation, gettingA' = 250 - (5/2) 2x= 250 - 5x= 5 (50 -x)= 0forx=50 .Note that since there are 5 lengths ofxin this construction and 500 feet of fencing, it follows that. See the adjoining sign chart forA' .

Ifx=50 ft. andy=125 ft. ,thenA= 6250 ft.2is the largest possible area of the pen.

PROBLEM 3:An open rectangular box with square base is to be made from 48 ft.2of material. What dimensions will result in a box with the largest possible volume ?

SOLUTION 3:Let variablexbe the length of one edge of the square base and variableythe height of the box.

The total surface area of the box is given to be48 = (area of base) + 4 (area of one side)=x2+ 4 (xy) ,so that4xy= 48 -x2or

.We wish to MAXIMIZE the total VOLUME of the boxV= (length) (width) (height)= (x) (x) (y) =x2y.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygettingV=x2y

= 12x- (1/4)x3.Now differentiate this equation, gettingV' = 12 - (1/4)3x2= 12 - (3/4)x2= (3/4)(16 -x2)= (3/4)(4 -x)(4 +x)= 0forx=4 orx=-4 .Butsince variablexmeasures a distance andx> 0 . Since the base of the box is square and there are 48 ft.2of material, it follows that. See the adjoining sign chart forV' .

Ifx=4 ft. andy=2 ft. ,thenV= 32 ft.3is the largest possible volume of the box.

PROBLEM 4:A container in the shape of a right circular cylinder with no top has surface area 3ft.2What height h and base radius r will maximize the volume of the cylinder ?

SOLUTION 4:Let variablerbe the radius of the circular base and variablehthe height of the cylinder.

The total surface area of the cylinder is given to be(area of base) + (area of the curved side),so that

or

.We wish to MAXIMIZE the total VOLUME of the cylinderV= (area of base) (height).However, before we differentiate the right-hand side, we will write it as a function ofronly. Substitute forhgetting

.Now differentiate this equation, getting

= 0forr=1 orr=-1 .Butsince variablermeasures a distance andr> 0 . Since the base of the box is a circle and there areft.2of material, it follows that. See the adjoining sign chart forV' .

Ifr=1 ft. andh=1 ft. ,thenft.3is the largest possible volume of the cylinder.

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

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

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

We wish to MAXIMIZE the total VOLUME of the boxV= (length) (width) (height)= (4-2x) (3-2x) (x) .Now differentiate this equation using the triple product rule, gettingV' = (-2) (3-2x) (x) + (4-2x) (-2) (x) + (4-2x) (3-2x) (1)= -6x+ 4x2- 8x+ 4x2+ 4x2- 14x+ 12= 12x2- 28x+ 12= 4 ( 3x2- 7x+ 3 )= 0for (Use the quadratic formula.),i.e., foror.Butsince variablexmeasures a distance. In addition, the short edge of the cardboard is 3 ft., so it follows that. See the adjoining sign chart forV' .

Ifft. ,thenft.3is largest possible volume of the box.

PROBLEM 6:Consider all triangles formed by lines passing through the point (8/9, 3) and both the x- and y-axes. Find the dimensions of the triangle with the shortest hypotenuse.

SOLUTION 6:Let variablexbe the x-intercept and variableythe y-intercept of the line passing throught the point (8/9, 3) .

Set up a relationship betweenxandyusing similar triangles.

One relationship is,so that.We wish to MINIMIZE the length of the HYPOTENUSE of the triangle.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygetting

.Now differentiate this equation using the chain rule and quotient rule, getting

(Factor a 2 out of the big brackets and simplify.)

= 0 ,so that (If, thenA=0 .).By factoring outx, it follows that,so that (IfAB= 0 , thenA=0 orB=0 .)x=0(Impossible, sincex> 8/9. Why ?) or.Then,so that(x-8/9)3= 8 ,x-8/9 = 2 ,andx= 26/9 .See the adjoining sign chart forH' .

Ifx= 26/9 andy=13/3 ,then

is the shortest possible hypotenuse.

PROBLEM 7:Find the point (x, y) on the graph ofnearest the point (4, 0).SOLUTION 7:Let (x,y) represent a randomly chosen point on the graph of.

We wish to MINIMIZE the DISTANCE between points (x,y) and (4, 0) ,

.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygetting

.Now differentiate this equation using the chain rule, getting

= 0 ,so that (If, thenA=0 .)2x-7 = 0 ,orx=7/2 .See the adjoining sign chart forL' .

Ifx= 7/2 and,then

is the shortest possible distance from (4, 0) to the graph of.

PROBLEM 8:A cylindrical can is to hold 20m.3The material for the top and bottom costs $10/m.2and material for the side costs $8/m.2Find the radius r and height h of the most economical can.SOLUTION 8:Let variablerbe the radius of the circular base and variablehthe height of the cylinder.

The total volume of the cylinder is given to be(area of base) (height),so that

.We wish to MINIMIZE the total COST of construction of the cylinderC= (total cost of bottom) + (total cost of top) + (total cost of side)= (unit cost of bottom)(area of bottom) + (unit cost of top)(area of top) + (unit cost of side) (area of side)

(For convenience drop thesigns until the end of the problem.).However, before we differentiate the right-hand side, we will write it as a function ofronly. Substitute forhgetting


(Get a common denominator and combine fractions.)

= 0 ,so that (If, thenA=0 .),r3= 8 ,orr= 2 .Since variablermeasures a distance, it must satisfyr> 0 . See the adjoining sign chart forC' .

Ifr=2 m. andh=5 m. ,then

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

SOLUTION 9:Let variablexbe the distance denoted in the given diagram.

Assume that you travel at the following rates :SWIM : 2 mphWALK : 3 mph .Recall that if travel is at a CONSTANT rate of speed, then(distance traveled) = (rate of travel) (time elapsed)orD=R T,so that time elapsed is.We wish to MINIMIZE the total TIME elapsedT= (swim time) + (walk time)= (swim distance)/(swim rate) + (walk distance)/(walk rate)


= 0 ,so that

and.Square both sides of this equation, getting9x2= 4 (1 +x2) = 4 + 4x2,so that5x2= 4 ,x2= 4/5 ,or.Butsince variablexmeasures a distance and. See the adjoining sign chart forT' .

Ifmi.thenhr.is the shortest possible time of travel.

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

SOLUTION 10:Let variablexbe the width and variableythe length of the rectangular portion of the window.

The semi-circular portion of the window has length(radius).The perimeter (distance around outside only) of the window is given to be

so that

or.We wish to MAXIMIZE the total AREA of the RECTANGLEA= (width) (length) =x y.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygettingA=x y


= 0for,i.e.,.Since variablexmeasures distance,. In addition,xis largest wheny= 0 and the window is in the shape of a semi-circle. Thus,(Why ?). See the adjoining sign chart forA' .

Ifft. and y=3 ft. ,thenft.2is the largest possible area of the rectangle.

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

SOLUTION 11:Let variablexbe the ADDITIONAL trees planted in the existing orchard. We wish to MAXIMIZE the total PRODUCTION of applesP= (number of trees) (apple output per tree)= ( 50 +x) ( 800 - 10x)= 40,000 + 300x- 10x2.Now differentiate this equation, gettingP' = 300 - 20x= 20 ( 15 -x)= 0forx=15 .See the adjoining sign chart forP' .

Ifx= 15 additional trees ,thenP= 42,250 applesis the largest possible production of apples.

PROBLEM 12:Find the dimensions of the rectangle of largest area which can be inscribed in the closed region bounded by the x-axis, y-axis, and graph ofy=8-x3. (See diagram.)

SOLUTION 12:Let variablexbe the length of the base and variableythe height of the inscribed rectangle.

We wish to MAXIMIZE the total AREA of the rectangleA= (length of base) (height) =xy.However, before we differentiate the right-hand side, we will write it as a function ofxonly. Substitute forygettingA=x y=x( 8 -x3)= 8x-x4.Now differentiate this equation, gettingA' = 8 - 4x3= 4 ( 2 -x3)= 0 ,so thatx3= 2and.Note that. See the adjoining sign chart forA' .

Ifandy= 6 ,then

is the largest possible area for the inscribed rectangle.

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

SOLUTION 13:Let variablerbe the length of the base and variablehthe height of the rectangle.

It is given that the perimeter of the rectangle is12 = 2r+ 2hso that2h= 12 - 2randh= 6 -r.We wish to MAXIMIZE the total VOLUME of the resulting CYLINDERV= (area of base) (height).However, before we differentiate the right-hand side, we will write it as a function ofronly. Substitute forhgetting


= 0forr=0 orr=4 .Since variablermeasures distance and the perimeter of the rectangle is 12,. See the adjoining sign chart forV' .

Ifr= 4 ft. andh= 2 ft. ,thenft.ft.3is the largest possible volume for the cylinder.PROBLEM 14:A movie screen on a wall is 20 feet high and 10 feet above the floor. At what distance x from the front of the room should you position yourself so that the viewing angleof the movie screen is as large as possible ? (See diagram.)

SOLUTION 14:Let variablebe the viewing angle and variablexthe distance as denoted in the diagram. We seek to write angleas a function of distancex. Introduce angleas in the diagram below.

It follows from basic trigonometry that

so that(Equation 1).In a similar fashion

so that,or(Equation 2).Usefrom Equation 1 to substitute into Equation 2, getting.We wish to MAXIMIZE angle THETA given in this equation. Differentiate this equation, getting

= 0 ,so that,30x2+ 3000 = 10x2+ 9000 ,20x2= 6000 ,x2= 300 ,for.Butsince variablexmeasures distance and. If, then

(These are well-known values from basic trigonometry.)

radiansordegrees .See the adjoining sign chart for.

Ifft.ft.thendegreesradiansis the largest possible viewing angle.

PROBLEM 15:Find the dimensions (radius r and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2.SOLUTION 15:Let variablerbe the radius of the circular base and variablehthe height of the inscribed cone as shown in the two-dimensional side view.

It is given that the circle's radius is 2. Find a relationship betweenrandh. Let variablezbe the height of the small right triangle.

By the Pythagorean Theorem it follows thatr2+z2= 22so thatz2= 4 -r2or.Thus the height of the inscribed cone is.We wish to MINIMIZE the total VOLUME of the CONE

PROBLEM 16:What anglebetween two edges of length 3 will result in an isosceles triangle with the largest area ? (See diagram.)

SOLUTION 16:Write the area of the given isosceles triangle as a function of. Let variablexbe the length of the base and variableythe height of the triangle, and consider angle. Write each ofxandyas functions of.

It follows from basic trigonometry that

so that(Equation 1 ),and

so that(Equation 2 )

We wish to MAXIMIZE the AREA of the isosceles triangleA= (1/2) (length of base) (height)= (1/2)xy.Before we differentiate, use Equations 1 and 2 to rewrite the right-hand side as a function ofonly. ThenA= (1/2)xy

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

(Factor out (9/2) and simplify the expression.)

= 0 ,so that

and.It follows algebraically (Why ?) that

so that from basic trigonometry we getor,and henceor.Becausemeasures an angle in a triangle, it is logical to assume that. Thus,. See the adjoining sign chart forA' .

Ifradians = 90 degrees,thenA= 9/2is the largest possible area for the triangle.PROBLEM 17:Of all lines tangent to the graph of, find the tangent lines of mimimum slope and maximum slope.

SOLUTION 17:We need to determine a general SLOPE EQUATION for tangent lines.

This means that we need the first derivative ofy. Differentiateusing the quotient rule, getting

.We wish to MAXIMIZE and MINIMIZE the SLOPE equation.Now differentiate this equation using the quotient rule and chain rule, getting

(Factor out -12 and (x2+3) from the numerator and simplify the expression.)

(Divide out a factor of (x2+3) .)

= 0 ,so that (If, thenA= 0 .)-36 ( 1 -x) ( 1 +x) = 0and (IfAB= 0 , thenA= 0 orB= 0 .)x=1 orx=-1 .See the adjoining sign chart forS' .

Ifx=-1 andy= 3/2 ,thenS= 3/4is the largest possible slope for this graph. The corresponding tangent line isy- 3/2 = 3/4(x- (-1) )ory= (3/4)x+ (9/4) .Ifx=1 andy= 3/2 ,thenS= -3/4is the smallest possible slope for this graph. The corresponding tangent line isy- 3/2 = -3/4(x- 1 )ory= (-3/4)x+ (9/4) .

PROBLEM 18:Find the length of the shortest ladder that will reach over an 8-ft. high fence to a large wall which is 3 ft. behind the fence. (See diagram.)

SOLUTION 18:Let variableLbe the length of the ladder resting on the top of the fence and touching the wall behind it. Let variablesxandybe the lengths as shown in the diagram.

WriteLas a function ofx. First find a relationship betweenyandxusing similar triangles. For example,

so that.We wish to MINIMIZE the LENGTH of the ladder.Before we differentiate, rewrite the right-hand side as a function ofxonly. Then


(Factor out 64xand (x-3) from the numerator of the fraction inside the brackets.)

(Divide out a factor of (x-3) and simplify the entire expression.)

(Factor out 2xfrom the numerator.)

= 0 ,so that (If, thenA= 0 .).Then (IfAB= 0 , thenA= 0 orB= 0 .)x=0or,so that,(x-3)3= 192 ,,and.Note thatx> 3 . See the adjoining sign chart forL' .

Ifft. andft. ,thenft.is the length of the shortest possible ladder.

PROBLEM 19:Find the point P = (x, 0) on the x-axis which minimizes the sum of the squares of the distances from P to (0, 0) and from P to (3, 2)..

SOLUTION 19:

Let variableSbe the sum of the squares of the distances between (0, 0) and (x, 0) ,,and between (3, 2) and (x, 0) ,.We wish to MINIMIZE the SUM of the squares of the distancesS=x2+ (x2-6x+13 ) = 2x2-6x+13 .Now differentiate, gettingS' = 4x-6= 4(x- 3/2)= 0forx= 3/2 .See the adjoining sign chart forS' .

Ifx= 3/2 ,thenS= 17/2is the smallest sum.

PROBLEM 20:Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same moment car A begins moving north at 60 mph. What will be the minimum distance between the cars and at what time t does the minimum distance occur ?SOLUTION 20:

Assume that the two cars travel at the following rates :CAR A : 60 mphCAR B : 90 mphLet variablexbe the distance car A travels inthours, and variableythe distance car B travels inthours. Let variableLbe the distance between cars A and B afterthours.

Thus, by the Pythagorean Theorem distanceLis.Before we differentiate, we will rewrite the right-hand side as a function oftonly. Recall that if travel is at a CONSTANT rate then(distance traveled) = (rate of travel) (time elapsed) .Thus, for car A the distance traveled afterthours is(Equation 1 )x= 60t,and for car B the distance traveled afterthours is(Equation 2 )y= 90t.Use Equations 1 and 2 to rewrite the equation forLas a function oftonly. Thus, we wish to MINIMIZE the DISTANCE between the two cars

.Differentiate, using the chain rule, getting

= 0so that (If, thenA= 0 .)23,400t- 5400 = 0 ,and.See the adjoining sign chart forL' .

Ifhrs. = 13.8 min. ,thenmi. ,mi. ,andmi.is the shortest possible distance between the cars.

PROBLEM 21:A rectangular piece of paper is 12 inches high and six inches wide. The lower right-hand corner is folded over so as to reach the leftmost edge of the paper (See diagram.).

Find the minimum length of the resulting crease.SOLUTION 21:Let variableLrepresent the length of the crease and let variablesxandybe as shown in the diagram.

We wish to writeLas a function ofx. Introduce variablewas shown in the following diagram.

It follows from the Pythagorean Theorem thatw2+ (6-x)2=x2,so thatw2=x2- (x2- 12x+ 36) = 12x- 36and.Find a relationship betweenxandy. The total area of the paper can be computed from the areas of three right triangles, two of which are exactly the same dimensions, and one rhombus. In particular72 = (total area of paper)= (area of small triangle) + 2(area of large triangle) + (area of rhombus)= (1/2)(length of base)(height) + 2(1/2)(length of base)(height) + (average height)(length of base)

,i.e.,.Solve this equation fory. Then,,,and.We wish to MINIMIZE the LENGTH of the crease.Before we differentiate, rewrite the right-hand side as a function ofxonly. Then


(Factor outxfrom the numerator.)

= 0 ,so that (If, thenA= 0 .).Thus, (IfAB= 0 , thenA= 0 orB= 0 .)x= 0or,,-2 (x-3)2= 3x- 18 ,-2 (x2- 6x+ 9) = 3x- 18 ,-2x2+ 12x-18 = 3x- 18 ,-2x2+ 9x= 0 ,x( -2x+ 9 ) = 0 ,so that (IfAB= 0 , thenA= 0 orB= 0 .)x= 0or( -2x+ 9 ) = 0 ,i.e.,x= 9/2 .Note that since the paper is 6 inches wide, it follows that. See the adjoining sign chart forL' .

Ifx= 9/2 in. andin.in. ,thenin.in.is the length of the shortest possible crease.

maximum and minimum

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