SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)

PROBLEM 1

What number exceeds its square by the maximum amount?

Solution:

Let

x = the number and

x2 = the square of the number

y = the difference between x and x2

y = x – x2

y' = 1 – 2x = 0

x = ½ answer

Problem 2

What positive number added to its reciprocal gives the minimum sum?

Solution:

Let

x = the required positive number and

1/x = the reciprocal of the number

y = sum of x and 1/x

y = x + 1/x

y = x + x-1

y' = 1 - x-2 = 0

x = 1 answer

Problem 3

The sum of two numbers is k. Find the minimum value of the sum of their squares.

Solution:

Let

x and y = the numbers

z = sum of their squares

k = x + y

y = k – x

z = x2 + y2

z = x2 + (k – x)2

dz/dx = 2x + 2(k - x)(-1) = 0

2x – k = 0

x = ½ k

y = k - ½ k

y = ½ k

z = (½ k)2 + (½ k)2

z = ½ k2

Problem 4

The sum of two numbers is k. Find the minimum value of the sum of their cubes.

Solution:

Let

x and y = the numbers

z = sum of their cubes

k = x + y

y = k – x

z = x3 + y3

z = x3 + (k – x)3

dz/dx = 3x2 + 3(k - x)2(-1) = 0

x2 – (k2 – 2kx + x2) = 0

x = ½ k

y = k - ½ k

y = ½ k

z = (½ k)3 + (½ k)3

z = ¼ k3

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 5

Let x and y = the numbers

x + y = 2 Equation (1)

1 + y’ = 0

y’ = –1

z = x3 + y2 Equation (2)

dz/dx = 3x2 + 2y y’ = 0

3x2 + 2y(–1) = 0

y = (3/2) x2

From Equation (1)

x + (3/2) x2 = 2

2x + 3x2 = 4

3x2 + 2x – 4 = 0

x = 0.8685 & –1.5352

use 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 = a

x = a – y

z = xy2

z = (a – y) y2

z = ay2 – y3

1

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)dz/dy = 2ay – 3y2 = 0

y = 2/3 a

x = a – 2/3 a

x = 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 = a

x = a – y

z = xy3

z = (a – y) y3

z = ay3 – y4

dz/dy = 3ay2 – 4y3 = 0

y2 (3a – 4y) = 0

y = 0 (absurd) and 3/4 a (use)

x = a - 3/4 a

x = 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 = a

1 + y’ = 0

y = –1

z = x2 y3

dz/dx = x2 (3y2 y’) + 2xy3 = 0

3x(–1) + 2y = 0

x = 2/3 y

2/3 y + y = a

5/3 y = a

y = 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:

A=xy

0=xy’+y

y’=–yx

Perimeter:

P=2x+2y

dPdx=2+2y’=0

1+(–yx)=0

y=x (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:

A=xy

0=xy’+y

y’=–yx

Perimeter:

P=x+2y

dPdx=1+2y’=0

1+2(–yx)=0

y=21x

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:

k=mx+n(2y+x)

k=mx+2ny+nx

k–(m+n)x=2ny

y=k2n−2nm+nx

Area:

A=xy

A=xk2n−2nm+nx

A=k2nx−2nm+nx2

dxdA=k2n−nm+nx=0

k2n=nm+nx

2

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)x=k2(m+n)

y=k2n−2nm+nk2(m+n)

y=k4n

A=k2(m+n)k4n

A=k28n(m+n) yard2 answer

Problem 12

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?

Solution

Area:

A=xy

0=xy’+y

y’=–yx

Fence:

P=2x+4y

dPdx=2+4y’=0

2+4(–yx)=0

y=21x

width = ½ × length answer

Problem 13

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

Solution

Area:

A=xy

0=xy’+y

y’=–yx

Fence:

P=2x+6y

dPdx=2+6y’=0

2+6(–yx)=0

y=31x

width =31 length 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.

Solution

Total cost:

300=2y+15(x–24)

y=168–075x

Area:

A=xy

A=x(168–075x)

A=168x–075x2

dxdA=168–15x=0

x=112ft

y=168–075(112)

y=84ft

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:

V=(9–2x)2x

V=81x–36x2+4x3

dxdV=81–72x+12x2=0

4x2–24x+27=0

using quadratic formula

a=4;b=–24;c=27

x=2a−bb2−4ac

x=2(4)−(−24)(−24)2−4(4)(27)

x=82412

x=45 and 15

use x = 1.5 inches

Maximum volume:

V=[9–2(15)]2(15)

V=54 in2 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:

V=(24–2x)(15–2x)x

V=360x–78x2+4x3

dxdV=360–156x+12x2=0

x2–13x+30=0

(x–10)(x–3)=0

x=10 (meaningless) and 3

Vmax=[24−2(3)][15–2(3)]3

Vmax=486 in3 answer

Problem 17

3

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)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

(a+b+a2−ab+b2)3b

Solution:

V=(6a–2x)(6b–2x)x

V=36abx–12(a+b)x2+4x3

dVdx=36ab–24(a+b)x+12x2=0

x2–2(a+b)x+3ab=0

A=1;B=–2(a+b);C=3ab

x=2A−BB2−4AC

x=2(1)2(a+b)4(a+b)2−4(1)(3ab)

x=22(a+b)2(a2+2ab+b2)−3ab

x=(a+b)+a2−ab+b2 and

x=(a+b)−a2−ab+b2

If a = b:

From

x=(a+b)+a2−ab+b2

x=(b+b)+b2−b2+b2

x=3b (x is equal to ½ of 6b - meaningless)

From

x=(a+b)−a2−ab+b2

x=(b+b)−b2−b2+b2

x=b ok

Use x=a+b−a2−ab+b2 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

b2+d2=D2

2bdddb+2d=0

dddb=−bd

Strength:

S=bd2

dddS=b(2d)+d2dddb=0

2bd+d2−bd=0

2bd=bd3

2b2=d2

d=2b

depth =2 breadth 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

b2+d2=D2

2bdddb+2d=0

dddb=−bd

Stiffness:

k=bd3

dddk=b(3d2)+d3dddb=0

3bd2+d3−bd=0

3bd2=bd4

3b2=d2

d=3b

depth =3 breadth 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"

4

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)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)

x2+y2=D2

2x+2yy’=0

y’=–xy

Perimeter

P=2x+2y

dPdx=2+2y’=0

2+2(–xy)=0

y=x

The largest rectangle is a square answer

See 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:

x2+y2=c2

2x+2yy’=0

y’=–xy

Area:

A=21xy

dAdb=21[xy’+y]=0

xy’+y=0

x(–xy)+y=0

y=x2y

y2=x2

y=x

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:

V=x(3x)yV=3x2y

0=3x2y’+6xy

y’=–2yx

Total Area:

AT=2(3x2)+2(3xy)+2(xy)

AT=6x2+8xy

dATdx=12x+8(xy’+y)=0

12x+8[x(–2yx)+y]=0

12x+8[–2y+y]=0

12x=8y

y=23x

altitude = 3/2 × shorter side of base answer

Problem 24

Solve Problem 23 if the box has an open top.

Solution:

Given Volume:

V=x(3x)y

V=3x2y

0=3x2y’+6xy

y’=–2yx

Area:

A=3x2+2(3xy)+2(xy)

A=3x2+8xy

dAdx=6x+8(xy’+y)=0

6x+8[x(–2yx)+y]=0

6x+8[–2y+y]=0

6x=8y

y=43x

altitude = 3/4 × shorter side of base answer

Problem 25

Find the most economical proportions of a quart can.

5

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)

Solution:

Volume:

V=41d2h= 1 quart

41d2dddh+2dh=0

dddh=−d2h

Total area (closed both ends):

AT=2(41d2)+dh

AT=21d2+dh

dddAT=d+ddddh+h=0

dddddh+h=0

d+d−d2h+h=0

d=h

diameter = height answer

Problem 26

Find the most economical proportions for a cylindrical cup.

Solution:

Volume:

V=41d2h

0=41d2dddh+2dh

dddh=−d2h

Area (open one end):

A=41d2+dh

dddA=21d+ddddh+h=0

21d+ddddh+h=0

21d+d−d2h+h=0

21d=h

But ½ d = radius, r

r = h

radius = height answer

Problem 27

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

Solution:

Volume:

V=x2y

0=x2y’+2xy

y’=–2yx

Area:

A=x2+4xy

dAdx=2x+4(xy’+y)=0

2x+4[x(–2yx)+y]=0

2x–8y+4y=0

2x=4y

x=2y

side of base = 2 × altitude answer

Problem 28

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

Solution:

Perimeter:

P=2y+x

0=2y’+1

y’=–12=–05

Area:

A=21xh

From the figure:

h=y2−025x2

A=21xy2−025x2

dxdA=21x2yy−05x2y2−025x2+y2−025x2=0

2y2−025x22(xyy−025x2)+y2−025x2=0

multiply both sides of the equation by y2−025x2

xyy–025x2+y2–025x2=0

xy(–05)–05x2+y2=0

y2–05xy–05x2=0

2y2–xy–x2=0

Solving for y by quadratic formula:

a = 2; b = –x; c = –x2

y=2a−bb2−4ac

y=2(2)−(−x)x2−4(2)(−x2)

y=4xx2+8x2

y=4x3x

y=x and −21x

y = -½ x is absurd, thus use y = x

Therefore

P=2x+x=3x

x=y=31P

h=y2−025x2

h=(31P)2−41(31P)2

h=91P2−136P2

h=112P2

h=123P

6

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS) A=21xh

Amax=21(31P)(123P)

Amax=1123P2

Amax=112333P2

[math]Amax=363P2 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:

a=4x+y

y=a–4x

Volume

V=x2y

V=x2(a–4x)

V=ax2–4x3

dVdx=2ax–12x2=0

2x(a–6x)=0

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

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

y=a−4(61a)

y=31a

Vmax=(31a)2(31a)

Vmax=1108a3 in3 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:

D2=d2+h2

0=2d+2hdddh

dddh=−hd

Volume of cylinder:

V=41d2h

dhdV=4d2dddh+2dh=0

ddddh+2h=0

d−hd+2h=0

2h=hd2

d2=2h2

d=2h

diameter =2 height 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):

A=dh

dddA=ddddh+h=0

ddddh+h=0

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

d−dddh+h=0

h=hd2

h2=d2

d=h

diameter = height 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:

A=xy

From the figure:

ya−x=ab

y=ab(a−x)

A=xab(a−x)

A=bx−abx2

dxdA=b−a2bx=0

a2bx=b

x=21a

y=ab(a−21a)

y=21b

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

7

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)largest rectangular building that can be erected, facing the hypotenuse of the triangle.

Solution:

Area:

A=xy

By similar triangle:

cot=rx=8060

r=43x

cot=sx=6080

s=34x

r+y+s=100

43x+y+34x=100

y=100−1225x

Thus,

A=x(100−1225x)

A=100x−1225x2

dxdA=100−1250x

x=100(5012

x=24 feet

y=100–1225(24)

y=50 feet

dimensions: 50 ft × 24 ft answer

Problem 34

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

Solution:

Area:

A=xy

By similar triangle:

cot=rx=ab

r=abx

cot=sx=ba

s=bax

r+y+s=a2+b2

abx+y+bax=a2+b2

y+aba2+b2x=a2+b2

y=a2+b2−aba2+b2x

y=ababa2+b2−(a2+b2)x

y=aba2+b2ab−xa2+b2

Thus,

A=xaba2+b2ab−xa2+b2

A=aba2+b2abx−x2a2+b2

dxdA=aba2+b2ab−2xa2+b2=0

2xa2+b2=ab

x=ab2a2+b2

y=aba2+b2ab−ab2a2+b2a2+b2

y=aba2+b2ab−2ab

y=aba2+b22ab

y=2a2+b2

Dimensions:

ab2a2+b22a2+b2 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:

(x–2)(y–3)=24

y=24x−2+3

Page area:

A=xy

A=x24x−2+3

A=24xx−2+3x

dxdA=(x−2)2(x−2)24−24x(1)+3=0

−48(x−2)2+3=0

–48+3(x–2)2=0

x=483+2

x=6 in

y=246−2+3

y=9 in

dimensions: 6 in × 9 in answer

Problem 36

8

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)A Norman window consists of a rectangle surmounted by a semicircle. What shape gives the most light for the given perimeter?

Solution:

Given perimeter:

P=b+2(h–r)+r

P=b+2h–2r+r

Where:

b = 2r[/math]

thus, r = ½ b

P=b+2h–b+21b

P=2h+21b

0=2dbdh+21

dbdh=−41

Light is most if area is maximum:

A=21r2+b(h–r)

A=21(21b)2+b(h–21b)

A=81b2+bh–21b2

A=81(−4)b2+bh

dbdA=82(−4)b+bdbdh+h=0

41b−b−41b+h=0

h=b

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

dbdh=−41

r=21b

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

A=41r2+b(h–r)

A=41(21b)2+b(h–21b)

A=116b2+bh–21b2

A=116(−8)b2+bh

dbdA=216(−8)b+bdbdh+h=0

81b−b−41b+h=0

−(1+81)b+h=0

h=(1+81)b

height=(1+81)breadth 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:

V=r2h

0=r2drdh+2rh

drdh=−r22rh

drdh=−r2h

Let

m = price per unit area of glass

½ m = price per unit area of plastic

k = total material cost per jar

k=m(2rh+r2)+21m(r2)

k=2mrh+23mr2

drdk=2mrdrdh+h+3mr=0

2rdrdh+h+3r=0

2r−r2h+h+3r=0

2h=3r

h=23r

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:

h2+2b−a2=a2

h=a2−4(b−a)2

9

SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)h=214a2−(b−a)2

Capacity is maximum if area is maximum:

A=21(b+a)h

A=21(b+a)214a2−(b−a)2

A=41(b+a)4a2−(b−a)2 (take note that a is constant)

dbdA=41(b+a)−2(b−a)24a2−(b−a)2+4a2−(b−a)2=0

4a2−(b−a)2=b2−a24a2−(b−a)2

4a2−(b−a)2=b2−a2

4a2–b2+2ab–a2=b2–a2

2b2–2ab–4a2=0

b2–ab–2a2=0

(b+a)(b–2a)=0

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

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

use b = 2a answer

Problem 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:

h2+2T−a2=2L−a2

h=4(L−a)2−4(T−a)2

h=21(L−a)2−(T−a)2

h=21L2−2La+a2−T2+2Ta−a2

h=21L2−T2−2(L−T)a

Area:

A=21(a+T)h

A=21(a+T)21L2−T2−2(L−T)a

A=41(a+T)L2−T2−2(L−T)a (note that L and T are constant)

dadA=41(a+T)−2(L−T)2L2−T2−2(L−T)a+L2−T2−2(L−T)a=0

L2−T2−2(L−T)a=(a+T)(L−T)L2−T2−2(L−T)a

L2–T2–2(L–T)a=(a+T)(L–T)

L2–T2–2La+2Ta=La–Ta+TL–T2

L2–TL=3La–3Ta

3(L–T)a=L(L–T)

a=31L

base = 1/3 × length of strip answer

Problem 42

From 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:

sin45=h214−a

h=214−asin45

h=214−a12

h=2214−a

Area:

A=A1+2A2

A=ah+2(21h2)

A=a2214−a+2214−a2

A=7a2−a222+8(14−a)2

dadA=72−a2−414−a=0

72−a2−27+4a=0

41−12a=27−72

a=317 in 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:

D1=x2+62=x2+36

D2=(10−x)2+182=(10−x)2+324

Total Distance:

D=D1+D2

D=x2+36+(10−x)2+324

dxdD=2x2x2+36+2(10−x)(−1)2(10−x)2+324=0

xx2+36=10−x(10−x)2+324

x(10−x)2+324=(10−x)x2+36

x2[(10–x)2+324]=(10–x)2(x2+36)

x2(10–x)2+324x2=x2(10–x)2+36(10–x)2

324x2=36(100–20x+x2)

9x2=100–20x+x2

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SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)8x2+20x–100=0

2x2+5x–25=0

(2x–5)(x+5)=0

For 2x - 5 = 0; x = 5/2

For x + 5 = 0; x = -5 (meaningless)

Use x = 5/2 = 2.5 mi

D=252+36+(10−25)2+324

D=26 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:

L1=x2+82=x2+64

L2=(15−x)2+122=(15−x)2+144

Total length of wire:

L=L1+L2

L=x2+64+(15−x)2+144

dxdL=2x2x2+64+2(15−x)(−1)2(15−x)2+144=0

xx2+64=15−x(15−x)2+144

x(15−x)2+144=(15−x)x2+64

x2[(15–x)2+144]=(15–x)2(x2+64)

x2(15–x)2+144x2=x2(15–x)2+64(15–x)2

144x2=64(225–30x+x2)

9x2=4(225–30x+x2)

5x2+120x–900=0

x2+24x–180=0

(x+30)(x–6)=0

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:

S1=x2+a2

S2=(c−x)2+b2

Total distance traveled by light:

S=S1+S2

S=x2+a2+(c−x)2+b2

dxdS=2x2x2+a2+2(c−x)(−1)2(c−x)2+b2

xx2+a2=c−x(c−x)2+b2

x(c−x)2+b2=(c−x)x2+a2

x2[(c–x)2+b2]=(c–x)2(x2+a2)

x2(c–x)2+b2x2=x2(c–x)2+a2(c–x)2

b2x2=a2(c2–2cx+x2)

b2x2=a2c2–2a2cx+a2x2

(a2–b2)x2–2a2cx+a2c2=0

By Quadratic Formula:

A = a2 – b2; B = –2a2c; C = a2c2

x=2A−BB2−4AC

x=2(a2–b2)2a2c4a4c2−4(a2–b2)(a2c2)

x=2(a2–b2)2a2c4a4c2−4a4c2+4a2b2c2

x=2(a2–b2)2a2c2abc

x=ac(ab)(a−b)(a+b)

For

x=ac(a+b)(a−b)(a+b)

x=aca−b meaningless if a > b

For

x=ac(a−b)(a−b)(a+b)

x=aca+b ok

Use x=aca+b

when S is minimum:

c−x=c−aca+b

c−x=a+b(a+b)c−ac

c−x=a+bac+bc−ac

c−x=bca+b

tan=xa

tan=aaca+b

tan=ca+b

tan=bc−x

tan=bbca+b

tan=ca+b

tan α = tan β

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SOLVED PROBLEMS IN MAXIMA AND MINIMA (DIFFIRENTIAL CALCULUS)thus, 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:

x2a2−b2y2=1

For equilateral hyperbola, b = a.

x2a2−a2y2=1

x2–y2=a2

x2=a2+y2

Distance d:

d=(x−0)2+(y−k)2

d=x2+(y−k)2

d=a2+y2+(y−k)2

dydd=2y+2(y−k)2a2+y2+(y−k)2=0

2y+2(y−k)

2y=k

y=21k

y2=41k2

Nearest Distance:

d=a2+41k2+(21k−k)2

d=a2+21k2 answer

Problem 47

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

Solution:

d=(x−4a)2+(y−0)2

d=(x−4a)2+y2

from

a2y=x3

y=x3a2

y2=x6a4

d=(x−4a)2+x6a4

dxdd=2(x−4a)+a46x5/2(x−4a)2+x6a4=0

(x−4a)+a43x5=0

3x5+a4x–4a5=0

by trial and error: x=a

y=a2a3=a

the nearest point is (a, a) answer

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