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