Problem 5

The 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 5Let x and y = the numbers

x + y = 2            Equation (1)1 + y' = 0y' = -1

z = x3 + y2            Equation (2)dz/dx = 3x2 + 2y y' = 03x2 + 2y(-1) = 0y = (3/2) x2

From Equation (1)x + (3/2) x2 = 22x + 3x2 = 43x2 + 2x - 4 = 0x = 0.8685 & -1.5352use x = 0.8685

y = (3/2)(0.8685)2

y = 1.1315

z = 0.86853 + 1.13152

z = 1.9354 answer

 Problem 6

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

Solution:Let x and y = the numbers

x + y = ax = a - y

z = xy2

z = (a - y) y2

z = ay2 - y3

dz/dy = 2ay - 3y2 = 0y = 2/3 a

x = a - 2/3 ax = 1/3 a

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

 

Problem 7

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

Solution:Let x and y the numbers

x + y = ax = a - y

z = xy3

z = (a - y) y3

z = ay3 - y4

dz/dy = 3ay2 - 4y3 = 0y2 (3a - 4y) = 0y = 0 (absurd) and 3/4 a (use)

x = a - 3/4 ax = 1/4 a

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

 Problem 8

Find 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:Let x and y the numbers

x + y = a1 + y' = 0y = -1

z = x2 y3

dz/dx = x2 (3y2 y') + 2xy3 = 03x(-1) + 2y = 0x = 2/3 y

2/3 y + y = a5/3 y = ay = 3/5 a

x = 2/3 (3/5 a)x = 2/5 a

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

Problem 9

What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fencing?

 

Solution:Area:

 

Perimeter:

(a square) answer

 Problem 10

A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing?

 Solution:Area:

 

Perimeter:

 

width = ½ × length answer

 

Problem 11

A rectangular lot is to be fenced off along a highway. If the fence on the highway costs m dollars per yard, on the other sides n dollars per yard, find the area of the largest lot that can be fenced off for k dollars.

 

Solution:Total cost:

 Area:

 

 

answer

A 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?

 

SolutionArea:

 

Fence:

 

width = ½ × length        answer 

Problem 13

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

SolutionArea:

 

Fence:

 

answer Problem 14

A 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. 

SolutionTotal cost:

 

Area:

 

 

Dimensions: 84 ft × 112 ft answer

Problem 15

A box is to be made of a piece of cardboard 9 inches square by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way.

 Solution:

 

 using quadratic formula

 use x = 1.5 inches

 Maximum volume:

answer

 Problem 16

Find the volume of the largest box that can be made by cutting equal squares out of the corners of a piece of cardboard of dimensions 15 inches by 24 inches, and then turning up the sides.

 Solution:

 

 

answer

 Problem 17

Find the depth of the largest box that can be made by cutting equal squares of side x out of the corners of a piece of cardboard of dimensions 6a, 6b, (b ≤ a), and then turning up the sides. To select that value of x which yields a maximum volume, show that

 

 Solution:

 

 

and

 If a = b:From

(x is equal to ½ of 6b - meaningless)

From

ok

 

Use answer

Problem 18

The strength of a rectangular beam is proportional to the breadth and the square of the depth. Find the shape of the largest beam that can be cut from a log of given size.

 

Solution:Diameter is given (log of given size), thus D is constant

 Strength:

             answer

 Problem 19

The stiffness of a rectangular beam is proportional to the breadth and the cube of the depth. Find the shape of the stiffest beam that can be cut from a log of given size.

 Solution:Diameter is given (log of given size), thus D is constant

 Stiffness:

             answer

 Problem 20

Compare for strength and stiffness both edgewise and sidewise thrust, two beams of equal length, 2 inches by 8 inches and the other 4 inches by 6 inches (See Problem 18 and Problem 19 above). Which shape is more often used for floor joist? Why?

 

Solution:Strength, S = bd2

Stiffness, k = bd3

 For 2" × 8":

Oriented such that the breadth is 2"S = 8(22) = 32 in3

k = 8(23) = 64 in4

Oriented such that the breadth is 8"S = 2(82) = 128 in3

k = 2(83) = 1024 in4

 For 4" × 6":

Oriented such that the breadth is 6"S = 6(42) = 96 in3

k = 6(43) = 384 in4

Oriented such that the breadth is 4"S = 4(62) = 144 in3

k = 4(63) = 864 in4

Strength wise and stiffness wise, 4" × 6" is more preferable. Economic wise, 2" × 8" is more preferable.

Problem 21

Find the rectangle of maximum perimeter inscribed in a given circle.

 Solution:Diameter D is constant (circle is given)

Perimeter

The largest rectangle is a square answerSee also the solution using trigonometric function.

 Problem 22

If the hypotenuse of the right triangle is given, show that the area is maximum when the triangle is isosceles.

 Solution:

 Area:

The triangle is an isosceles right triangle answer

 Problem 23

Find the most economical proportions for a covered box of fixed volume whose base is a rectangle with one side three times as long as the other.

 Solution:Given Volume:

 Total Area:

altitude = 3/2 × shorter side of base answer

 Problem 24

Solve Problem 23 if the box has an open top.

 Solution:Given Volume:

 Area:

altitude = 3/4 × shorter side of base answer

Problem 25

Find the most economical proportions of a quart can.

 Solution:Volume:

 Total area (closed both ends):

diameter = height            answer

 Problem 26

Find the most economical proportions for a cylindrical cup.

 Solution:Volume:

 Area (open one end):

 But ½ d = radius, rr = hradius = height           answer

 Problem 27

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

 Solution:Volume:

 

Area:

side of base = 2 × altitude            answer

Problem 28

The perimeter of an isosceles triangle is P inches. Find the maximum area.

 Solution:Perimeter:

 

Area:

From the figure:

 

multiply both sides of the equation by

 Solving for y by quadratic formula:a = 2; b = -x; c = -x2

 y = -½ x is absurd, thus use y = x

 Therefore

 

 

           answer

 Problem 29

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

Solution:

 

Volume

 

 

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

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

Use x = 1/6 a 

 

          answer 

Problem 30

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

Solution:

 From the figure:

 

Volume of cylinder:

 

          answer 

Problem 31

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

Solution:Convex surface area (shaded area):

 

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

          answer

Problem 32

Find the dimension of the largest rectangular building that can be placed on a right-triangular lot, facing one of the perpendicular sides.

 Solution:Area:

 

From the figure:

 

 

 

dimensions: ½ a    ×    ½ b            answer

 Problem 33

A 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 the triangle.

 Solution:Area:

 By similar triangle:

 

 Thus,

 

 dimensions: 50 ft × 24 ft            answer

 Problem 34

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

 Solution:Area:

 

By similar triangle:

 

 

Thus,

 

 

Dimensions:

           answer

Problem 35

A page is to contain 24 sq. in. of print. The margins at top and bottom are 1.5 in., at the sides 1 in. Find the most economical dimensions of the page. 

Solution:Print Area:

 

Page area:

 

 dimensions:    6 in    ×    9 in            answer 

Problem 36

A Norman window consists of a rectangle surmounted by a semicircle. What shape gives the most light for the given perimeter? 

Solution:Given perimeter:

Where:

thus, r = ½ b

 

Light is most if area is maximum:

breadth = height            answer 

Problem 37

Solve Problem 36 above if the semicircle is stained glass admitting only half the normal amount of light. 

Solution:From Solution of Problem 36

 

Half amount of light is equivalent to half of the area.

           answer

Problem 38

A cylindrical glass jar has a plastic top. If the plastic is half as expensive as glass, per unit area, find the most economical proportion of the jar.

 

Solution:Volume:

Letm = price per unit area of glass½ m = price per unit area of plastick = total material cost per jar

height = 3/2 × radius of base             answer

 

Problem 39

A trapezoidal gutter is to be made from a strip of tin by bending up the edges. If the cross-section has the form

shown in Fig. 38, what width across the top gives maximum carrying capacity?

 Solution:

 

 

Capacity is maximum if area is maximum:

           (take note that a is constant)

 

For b + a = 0; b = -a (meaningless)For b - 2a = 0; b = 2a (ok)

use b = 2a            answer

Problem 43

A ship lies 6 miles from shore, and opposite a point 10 miles farther along the shore another ship lies 18 miles offshore. A boat from the first ship is to land a passenger and then proceed to the other ship. What is the least distance the boat can travel?

 Solution:

 

 

Total Distance:

 For 2x - 5 = 0; x = 5/2For x + 5 = 0; x = -5 (meaningless)Use x = 5/2 = 2.5 mi

            answer

 Problem 44

Two posts, one 8 feet high and the other 12 feet high, stand 15 ft apart. They are to be supported by wires attached to a single stake at ground level. The wires running to the tops of the posts. Where should the stake be placed, to use the least amount of wire?

 Solution:

Total length of wire:

 For x + 30 = 0; x = -30 (meaningless)For x - 6 = 0; x = 6 (ok)use x = 6 ft

 location of stake: 6 ft from the shorter post            answer

 Problem 45

A ray of light travels, as in Fig. 39, from A to B

via the point P on the mirror CD. Prove that the length (AP + PB) will be a minimum if and only if α = β.

 Solution:

 

Total distance traveled by light:

 

By Quadratic Formula:A = a2 - b2; B = -2a2c; C = a2c2

 

For

      meaningless if a > b

 

For

      ok

 

Use

 

when S is minimum:

 

 

 

tan α = tan βthus, α = β            (ok!)

Problem 46

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

 

Solution:Standard equation:

For equilateral hyperbola, b = a.

Distance d:

 

Nearest Distance:

           answer

 

Problem 47

Find 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

Problem 48

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

= x3.

 

Solution:

 

from

 

For 3x + 10 = 0; x = -10/3 (meaningless)For x - 2 = 0; x = 2 (ok)use x = 2

            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 (ok)

 

Problem 49

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

 

Solution:

 

From

 

y = -8/3 a is meaningless, use y = 2a

 

           answer

 

Problem 50

Find 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 (ok)

 

Problem 51

Find 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 52

Find 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 54

A 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 55

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

 

Solution:

 

Volume

 

height = (m + 1) × radius            answer

Problem 56

The 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:Letk = unit price of pinemk = unit price of oakC = total cost

 

Volume of the square box:

 

Total cost:

           answer

 

Problem 57

A 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:Letk = unit price

 

Total cost:

 

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

 

total height = diameter            answer

Problem 58

For 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:Letk = unit price of wall2k = unit price of floor3k = unit price of roof

 

Total cost:

 

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

 

diameter = 2/7 × total height            answer

 

Problem 59

An 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:Area of the floor

 

Area of cylindrical wall

 

Area of conical roof:

 

Total area:

 

Volume = volume of cylinder + volume of cone

 

height of cone = 2 × height of cylinder            answer

Problem 60

One corner of a leaf of width a is folded over so as just to reach the opposite side of the page. Find the width of the part folded over when the length of the crease is a minimum. See Figure 41.

 

Solution:From the figure:

 

By double angle formula: sin 2α = cos2

α - sin2 α:

           answer

 

Problem 61

Solve Problem 60 above if the area folded over is to be a minimum.

 

Solution:From the solution of Problem 60 above:

 

Thus,

 

Area:

           answer

Problem 62

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

 

Solution:By 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 63

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

 

Solution:Volume 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 64

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

 Solution:Volume of sphere or radius a:

 Volume of cone of radius r and altitude h:

 

From the solution of Problem 63:

 

 

           answer

 

Problem 65

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

 Solution.Volume of cone:

 

By similar triangle:

 

Thus,

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

 

Another Solution:For 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,

 

           (ok!)

66 - 68 Maxima and minima: Pyramid inscribed in a sphere and Indian tepee

A D V E R T I S E M E N T

Problem 66

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

 

Solution:

Volume of pyramid:

 

From the figure:

 

 

 

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

 Problem 67

An 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:

r^2 = 2h

           answer

 

Problem 68

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

 

Solution:Area of covering material:

where

 

Volume of tepee:

           answer

Problem 69

A 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

0.06 per mile, how much must he pay for the trip?

 

Solution:

Distance traveled by boat:

 

Note: time = distance/speed

 

Total cost of travel:

 answer

 

Problem 70

A 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:distance = speed × time

 

           answer

 Problem 71

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

 Solution:By Pythagorean Theorem:

 

           answer

Problem 72

A 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 73

It 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:Attraction:

           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