Maxima and Minima Problems.docx

Problem 5The sum of two positive numbers is 2. Find the smallest value possible for the sum of the cube of one number and the square of the other.

Solution 5

Click here to show or hide the solutionLet x and y = the numbers

→ Equation (1)

→ Equation (2)

From Equation (1)

Use

answer

Problem 6Find two numbers whose sum is a, if the product of one to the square of the other is to be a minimum.

Solution:

Click here to show or hide the solutionLet x and y = the numbers

The numbers are 1/3 a, and 2/3 a. answer

Problem 7Find two numbers whose sum is a, if the product of one by the cube of the other is to be a maximum.

Solution:

Click here to show or hide the solutionLet x and y the numbers

The numbers are 1/4 a and 3/4 a. answer

Problem 8Find two numbers whose sum is a, if the product of the square of one by the cube of the other is to be a maximum.

Solution:

Click here to show or hide the solutionLet x and y the numbers

The numbers are 2/5 a and 3/5 a. answer

Problem 12A rectangular field of fixed area is to be enclosed and divided into three lots by parallels to one of the sides. What should be the relative dimensions of the field to make the amount of fencing minimum?

Solution

Click here to show or hide the solutionArea:

Fence:

width = ½ × length answer

Problem 13Do Ex. 12 with the words "three lots" replaced by "five lots".

Solution


Fence:

answer

Problem 14A rectangular lot is bounded at the back by a river. No fence is needed along the river and there is to be 24-ft opening in front. If the fence along the front costs $1.50 per foot, along the sides $1 per foot, find the dimensions of the largest lot which can be thus fenced in for $300.

Solution

Click here to show or hide the solutionTotal cost:

Area:

Dimensions: 84 ft × 112 ft answer

Solution:

Click here to show or hide the solutionVolume:

Total area (closed both ends):

Diameter = height answer

Problem 26Find the most economical proportions for a cylindrical cup.

Solution:


Area (open one end):

Radius = height answer

Problem 27Find the most economical proportions for a box with an open top and a square base.

Solution:


Area:

Aide of base = 2 × altitude answer

29 - 31 Solved problems in maxima and minimaSubmitted by Romel Verterra on May 26, 2009 - 10:40am

Problem 29The sum of the length and girth of a container of square cross section is a inches. Find the maximum

volume.

Solution:

Click here to show or hide the solution

Volume

For 2x = 0; x = 0 (meaningless)

For a - 6x = 0; x = 1/6 a

Use x = 1/6 a

answer

Problem 30Find the proportion of the circular cylinder of largest volume that can be inscribed in a given sphere.

Solution:

Click here to show or hide the solutionFrom the figure:

Volume of cylinder:

answer

Problem 31In Problem 30 above, find the shape of the circular cylinder if its convex surface area is to be a maximum.

Solution:

Click here to show or hide the solutionConvex surface area (shaded area):

From Solution to Problem 30 above, dh/dd = -d/h

answer

Problem 32Find the dimension of the largest rectangular building that can be placed on a right-triangular lot, facing

one of the perpendicular sides.

Solution:


From the figure:

Dimensions: ½ a × ½ b answer

Problem 33A lot has the form of a right triangle, with perpendicular sides 60 and 80 feet long. Find the length and width of the largest rectangular building that can be erected, facing the hypotenuse of thetriangle.

Solution:


By similar triangle:

Thus,

Dimensions: 50 ft × 24 ft answer

Problem 34Solve Problem 34 above if the lengths of the perpendicular sides are a, b.

Solution:



Thus,

Dimensions:

answer

46 - 47 Solved Problems in Maxima and MinimaSubmitted by Romel Verterra on June 5, 2009 - 8:36am

Problem 46Given point on the conjugate axis of an equilateral hyperbola, find the shortest distance to the curve.

Solution:

Click here to show or hide the solutionStandard equation:

For equilateral hyperbola, b = a.

Thus,

Distance d:

Nearest Distance:

answer

Problem 47Find the point on the curve a2 y = x3 that is nearest the point (4a, 0).

Solution:


from

by trial and error:

The nearest point is (a, a). answer

41 - 42 Maxima and Minima Problems Involving Trapezoidal Gutter

Submitted by Romel Verterra on June 2, 2009 - 4:10pmProblem 41

In Problem 39, if the strip is L in. wide, and the width across the top is T in. (T < L), what base width gives the maximum capacity?

Solution:


Area:

(note that L and T are constant)

Base = 1/3 × length of strip answer

Problem 42From a strip of tin 14 inches a trapezoidal gutter is to be made by bending up the sides at an angle of 45°. Find the width of the base for greatest carrying capacity.

Solution:


Area:

answer

48 - 49 Shortest distance from a point to a curve by maxima and minima

Submitted by Romel Verterra on June 5, 2009 - 9:45amProblem 48

Find the shortest distance from the point (5, 0) to the curve 2y2 = x3.

Solution:


from

For , (meaningless)

For , (okay)

Use .

answer

Another Solution:


Differentiate

→ slope of tangent at any point

Thus, the slope of normal at any point is

Equation of normal:

the same equation as above (okay)

Problem 49Find the shortest distance from the point (0, 8a) to the curve ax2 = y3.

Solution:


From

is meaningless, use

answer

Problem 50Find the shortest distance from the point (4, 2) to the ellipse x2 + 3y2 = 12.

Solution:


from

By trial and error

The nearest point is (3, 1)

Nearest distance:

answer

Another Solution:


→ slope of tangent at any point

Thus, slope of normal at any point is

Equation of normal:

the same equation as above (okay)

Problem 51Find the shortest distance from the point (1 + n, 0) to the curve y = xn, n > 0.

Solution:


by inspection: x = 1

1 raise to any positive number is 1

answer

Problem 52Find the shortest distance from the point (0, 5) to the ellipse 3y2 = x3.

Solution:


slope of tangent at any point

Thus, slope of normal at any point is

Equation of normal:

By trial and error

Nearest point on the curve is (3, 3)

Shortest distance

answer

Problem 53Cut the largest possible rectangle from a circular quadrant, as shown in Fig. 40.

Solution:


Area of rectangle

for

(meaningless)

for

answer

Problem 54A cylindrical tin boiler, open at the top, has a copper bottom. If sheet copper is m times as expensive as tin, per unit area, find the most economical proportions.

Solution:


Letk = cost per unit area of tinmk = cost per unit area of copperC = total cost

Volume

Height = m × radius answer

Problem 55Solve Problem 54 above if the boiler is to have a tin cover. Deduce the answer directly from thesolution of Problem 54.

Solution:


Volume

Height = (m + 1) × radius answer

Problem 72A light is to be placed above the center of a circular area of radius a. What height gives the best illumination on a circular walk surrounding the area? (When light from a point source strikes a surface obliquely, the intensity of illumination is

where θ is the angle of incidence and d the distance from the source.)

Solution:


From the figure:

Thus,

answer

Problem 73It is shown in the theory of attraction that a wire bent in the form of a circle of radius a exerts upon a particle in the axis of the circle (i.e., in the line through the center of the circle perpendicular to the plane) an attraction proportional to

where h is the height of the particle above the plane of the circle. Find h, for maximum attraction. (Compare with Problem 72 above)

Solution:

Click here to show or hide the solutionAttraction:

answer

Problem 74In Problem 73 above, if the wire has instead the form of a square of side , the attraction is proportional to

Find h for maximum attraction.

Solution:


Use

answer

ProblemFrom the right triangle ABC shown below, AB = 40 cm and BC = 30 cm. Points E and F areprojections of point D from hypotenuse AC to the perpendicular legs AB and BC, respectively. How far is D from AB so that length EF is minimal?

Solution

Click here to show or hide the solutionBy ratio and proportion

By Pythagorean theorem

For minimum length of d, differentiate then equate to zero

Distance of D from side AB for minimum length of d

answer

The same problem was solved by Geometry alone. See the solution here: Distance between projection points

] Problem 69A man on an island 12 miles south of a straight beach wishes to reach a point on shore 20 miles east. If a motorboat, making 20 miles per hour, can be hired at the rate of $2.00 per hour for the time it is actually used, and the cost of land transportation is $0.06 per mile, how much must he pay for the trip?

Solution:

Click here to show or hide the solutionDistance traveled by boat:

Note: time = distance/speed

Total cost of travel:

answer

Problem 70A man in a motorboat at A (Figure 42) receives a message at noon calling him to B. A bus making 40 miles per hour leaves C, bound for B, at 1:00 PM. If AC = 40 miles, what must be the speed of the boat to enable the man to catch the bus.

Solution:

Click here to show or hide the solutiondistance = speed × time

answer

Problem 71In Problem 70, if the speed of the boat is 30 miles per hour, what is the greatest distance offshorefrom which the bus can be caught?

Solution:

Click here to show or hide the solutionBy Pythagorean Theorem:

answer

Problem 66Find the largest right pyramid with a square base that can be inscribed in a sphere of radius a.

Solution:

Click here to show or hide the solutionVolume of pyramid:

From the figure:

Altitude of pyramid = 4/3 × radius of sphere, a answer

Problem 67An Indian tepee is made by stretching skins or birch bark over a group of poles tied together at the top. If poles of given length are to be used, what shape gives maximum volume?

Solution:


/>From the figure:

The length of pole is given, thus L is constant

Volume of tepee:

answer

Problem 68Solve Problem 67 above if poles of any length can be found, but only limited amount of covering material is available.

Solution:

Click here to show or hide the solutionArea of covering material:

where

Volume of tepee:

answer

Problem 62Inscribe a circular cylinder of maximum convex surface area in a given circular cone.

Solution:

Click here to show or hide the solutionBy similar triangle:

Convex surface area of the cylinder:

The cone is given, thus H and D are constant

Diameter of cylinder = radius of cone answer

Problem 63Find the circular cone of maximum volume inscribed in a sphere of radius a.

Solution:

Click here to show or hide the solutionVolume of the cone:

From the figure:

The sphere is given, thus radius a is constant.

Altitude of cone = 4/3 of radius of sphere answer

Problem 64A sphere is cut to the shape of a circular cone. How much of the material can be saved? (SeeProblem 63).

Solution:

Click here to show or hide the solutionVolume of sphere or radius a:

Volume of cone of radius r and altitude h:

From the solution of Problem 63:

Thus,

answer

Problem 65Find the circular cone of minimum volume circumscribed about a sphere of radius a.

Solution:

Click here to show or hide the solutionVolume of cone:


Thus,

Altitude of the cone = 4 × the radius of the sphere, a answer

Another Solution:

Click here to show or hide the solutionFor a circle inscribed in a triangle, its center is at the point of intersection of the angular bisector of the triangle called the incenter (see figure).

For the problem:

From the figure:

Thus,

(okay!)

Problem 58For the silo of Problem 57, find the most economical proportions, if the floor is twice as expensive as the walls, per unit area, and the roof is three times as expensive as the walls, per unit area.

Solution:

Click here to show or hide the solutionLetk = unit price of wall2k = unit price of floor3k = unit price of roof

Total cost:

→ Equation (1)

Volume of silo = volume of cylinder + volume of hemisphere:

→ Equation (2)

Equate Equations (1) and (2)

Diameter = 2/7 × total height answer

Problem 59An oil can consists of a cylinder surmounted by a cone. If the diameter of the cone is five-sixths of its height, find the most economical proportions.

Solution:

Click here to show or hide the solutionArea of the floor

Area of cylindrical wall

Area of conical roof:

Thus,

Total area:

→ Equation (1)

Volume = volume of cylinder + volume of cone

→ Equation (2)

Equate Equations (1) and (2)

Height of cone = 2 × height of cylinder answer

Problem 56The base of a covered box is a square. The bottom and back are made of pine, the remainder of oak. If oak is m times as expensive as pine, find the most economical proportion.

Solution:

Click here to show or hide the solutionLetk = unit price of pinemk = unit price of oakC = total cost

Volume of the square box:

Total cost:

answer

Problem 57A silo consists of a cylinder surmounted by a hemisphere. If the floor, walls, and roof are equally expensive per unit area, find the most economical proportion.

Solution:

Click here to show or hide the solutionLetk = unit price

Total cost:

Volume of silo = volume of cylinder + volume of hemisphere:

Total height = diameter answer

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Maxima and Minima Problems.docx

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