Maxima And Minima

c
 
 c       


c y=k²x

Problem 5

PROBLEM 1

z = x2 + y2

What number exceeds its square by the maximum amount?

z = x2 + (k ² x)2

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.

dz/dx = 2x + 2(k - x)(-1) = 0 Solution 5

Solution: 2x ² k = 0

Let x and y = the numbers

Let x=½k

x+y=2

x = the number and

Equation (1)

1 + y· = 0

x2 = the square of the number y=k-½k

y· = ²1

y = the difference between x and x2 y=½k

z = x3 + y2

Equation (2)

dz/dx = 3x2 + 2y y· = 0

y = x ² x2 z = (½ k)2 + (½ k)2

3x2 + 2y(²1) = 0

y' = 1 ² 2x = 0 z = ½ k2

y = (3/2) x2

x = ½ answer Problem 4

From Equation (1) Problem 2 What positive number added to its reciprocal gives the minimum sum?

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

x + (3/2) x2 = 2

Solution:

2x + 3x2 = 4

Let

3x2 + 2x ² 4 = 0

x and y = the numbers

x = 0.8685 & ²1.5352

z = sum of their cubes

use x = 0.8685

Solution: Let x = the required positive number and k=x+y 1/x = the reciprocal of the number y=k²x

y = (3/2)(0.8685)2

y = sum of x and 1/x y = 1.1315 y = x + 1/x z = x3 + y3 y = x + x-1 z = x3 + (k ² x)3

z = 0.86853 + 1.13152

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

z = 1.9354 answer

x2 ² (k2 ² 2kx + x2) = 0

Problem 6

x=½k y=k-½k

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

y=½k

Solution:

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.

Let x and y = the numbers Solution: z = (½ k)3 + (½ k)3

x+y=a

z = ¼ k3

x=a²y

Let x and y = the numbers z = sum of their squares z = xy2 k=x+y

Y Y Y

c
 
 c       


c z = (a ² y) y2

1 + y· = 0

Problem 10

z = ay2 ² y3

y = ²1

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?

dz/dy = 2ay ² 3y2 = 0 y = 2/3 a

z = x2 y3 dz/dx = x2 (3y2 y·) + 2xy3 = 0

x = a ² 2/3 a

3x(²1) + 2y = 0

x = 1/3 a

x = 2/3 y

Solution: Area:

The numbers are 1/3 a, and 2/3 a. answer Problem 7

2/3 y + y = a

A=xy

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

5/3 y = a

0=xy·+y

y = 3/5 a

y·=²yx

Solution: Let x and y the numbers

x = 2/3 (3/5 a)

x+y=a

x = 2/5 a

Perimeter:

x=a²y

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

P=x+2y

Problem 9

dPdx=1+2y·=0

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?

1+2(²yx)=0

z = xy3 z = (a ² y) y3

y=21x width = ½ × length answer

z = ay3 ² y4 Solution:

Problem 11

dz/dy = 3ay2 ² 4y3 = 0 Area:

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.

y2 (3a ² 4y) = 0 A=xy y = 0 (absurd) and 3/4 a (use) 0=xy·+y y·=²yx x = a - 3/4 a

Solution: x = 1/4 a Total cost: Perimeter:

k=mx+n(2y+x)

P=2x+2y

k=mx+2ny+nx

dPdx=2+2y·=0

k²(m+n)x=2ny

1+(²yx)=0

y=k2nï2nm+nx

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.

y=x (a square) answer

Area:

Solution: A=xy Let x and y the numbers

A=xk2nï2nm+nx x+y=a

Y
Y Y

c
 
 c       


c y·=²yx A=k2nxï2nm+nx2

Fence:

dxdA=k2nïnm+nx=0

P=2x+6y

k2n=nm+nx

dPdx=2+6y·=0

x=k2(m+n)

2+6(²yx)=0

y=k2nï2nm+nk2(m+n)

y=31x

y=k4n

width =31 length answer

Solution: V=(9²2x)2x V=81x²36x2+4x3 A=k2(m+n)k4n

Problem 14 dxdV=81²72x+12x2=0

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?

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.

4x2²24x+27=0

using quadratic formula a=4;b=²24;c=27 x=2aïbb2ï4ac

Solution Solution

Total cost:

Area:

300=2y+15(x²24)

A=xy

y=168²075x

x=2(4)ï(ï24)(ï24)2ï4(4)(27) x=82412 x=45 and 15 use x = 1.5 inches

0=xy·+y

Maximum volume:

y·=²yx

V=[9²2(15)]2(15)

Area: Fence:

A=xy

P=2x+4y

A=x(168²075x)

dPdx=2+4y·=0

A=168x²075x2

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.

2+4(²yx)=0 dxdA=168²15x=0

y=21x

x=112ft width = ½ × length

answer

y=168²075(112)

Problem 13

y=84ft

Do Ex. 12 with the words ´three lotsµ replaced by ´five lotsµ.

Dimensions: 84 ft × 112 ft answer Problem 15

Solution:

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.

V=(24²2x)(15²2x)x

Solution Area: A=xy 0=xy·+y

V=360x²78x2+4x3 dxdV=360²156x+12x2=0 x2²13x+30=0

Y Y Y

c
 
 c       


c (x²10)(x²3)=0

x=b ok

x=10 (meaningless) and 3 Vmax=[24ï2(3)][15²2(3)]3 Vmax=486 in3 answer

Use x=a+bïa2ïab+b2 answer

Problem 17

Problem 18

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

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

(a+b+a2ïab+b2)3b

2bdddb+2d=0 dddb=ïbd

Stiffness: k=bd3

Solution: V=(6a²2x)(6b²2x)x V=36abx²12(a+b)x2+4x3 dVdx=36ab²24(a+b)x+12x2=0 x2²2(a+b)x+3ab=0 A=1;B=²2(a+b);C=3ab x=2AïBB2ï4AC x=2(1)2(a+b)4(a+b)2ï4(1)(3ab) x=22(a+b)2(a2+2ab+b2)ï3ab x=(a+b)+a2ïab+b2 and x=(a+b)ïa2ïab+b2

dddk=b(3d2)+d3dddb=0 Solution: 3bd2+d3ïbd=0 Diameter is given (log of given size), thus D is constant

3bd2=bd4

b2+d2=D2

3b2=d2

2bdddb+2d=0

d=3b

dddb=ïbd

depth =3 breadth

Strength:

Problem 20

S=bd2

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?

dddS=b(2d)+d2dddb=0 2bd+d2ïbd=0 2bd=bd3

If a = b: From x=(a+b)+a2ïab+b2 x=(b+b)+b2ïb2+b2 x=3b (x is equal to ½ of 6b - meaningless) From x=(a+b)ïa2ïab+b2

answer

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.

x=(b+b)ïb2ïb2+b2

Y Y Y

c
 
 c       


c Solution:

Perimeter

Strength, S = bd2

P=2x+2y

Stiffness, k = bd3

dPdx=2+2y·=0

For 2" × 8":

2+2(²xy)=0

Oriented such that the breadth is 2"

y=x

Solution:

S = 8(22) = 32 in3

The largest rectangle is a square answer

Given Volume:

k = 8(23) = 64 in4

See also the solution using trigonometric function.

V=x(3x)yV=3x2y 0=3x2y·+6xy

Oriented such that the breadth is 8" Problem 22

y·=²2yx

S = 2(82) = 128 in3 k = 2(83) = 1024 in4

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

For 4" × 6":

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

Oriented such that the breadth is 6"

AT=6x2+8xy

S = 6(42) = 96 in3

dATdx=12x+8(xy·+y)=0

k = 6(43) = 384 in4

Solution:

12x+8[x(²2yx)+y]=0

Oriented such that the breadth is 4"

x2+y2=c2

12x+8[²2y+y]=0

S = 4(62) = 144 in3

2x+2yy·=0

12x=8y

k = 4(63) = 864 in4

y·=²xy

y=23x

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

Area:

altitude = 3/2 × shorter side of base answer

Problem 21

dAdb=21[xy·+y]=0

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

xy·+y=0

A=21xy Problem 24 Solve Problem 23 if the box has an open top.

x(²xy)+y=0 y=x2y y2=x2 y=x

Solution:

The triangle is an isosceles right triangle answer Solution:

V=x(3x)y

Problem 23 Diameter D is constant (circle is given) x2+y2=D2 2x+2yy·=0

Given Volume:

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.

y·=²xy

V=3x2y 0=3x2y·+6xy y·=²2yx Area: A=3x2+2(3xy)+2(xy) A=3x2+8xy

Y Y Y

c
 
 c       


c dAdx=6x+8(xy·+y)=0

Solution:

x=2y

6x+8[x(²2yx)+y]=0

Volume:

side of base = 2 × altitude

6x+8[²2y+y]=0

V=41d2h

Problem 28

6x=8y

0=41d2dddh+2dh

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

y=43x

dddh=ïd2h

altitude = 3/4 × shorter side of base answer

Area (open one end):

answer

A=41d2+dh Problem 25 dddA=21d+ddddh+h=0 Find the most economical proportions of a quart can.

21d+ddddh+h=0 Solution: 21d+dïd2h+h=0 Perimeter: 21d=h P=2y+x But ½ d = radius, r 0=2y·+1 r=h y·=²12=²05 radius = height

answer Area:

Solution:

Problem 27 A=21xh

Volume:

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

V=41d2h= 1 quart

h=y2ï025x2

41d2dddh+2dh=0

A=21xy2ï025x2

dddh=ïd2h

dxdA=21x2yyï05x2y2ï025x2+y2ï025x2=0

Total area (closed both ends):

2y2ï025x22(xyyï025x2)+y2ï025x2=0

AT=2(41d2)+dh

multiply both sides of the equation by y2ï025x2

AT=21d2+dh dddAT=d+ddddh+h=0 dddddh+h=0 d+dïd2h+h=0 d=h diameter = height

From the figure:

answer

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

Solution:

xyy²025x2+y2²025x2=0

Volume:

xy(²05)²05x2+y2=0

V=x2y

y2²05xy²05x2=0

0=x2y·+2xy

2y2²xy²x2=0

y·=²2yx

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

Area: A=x2+4xy

y=2aïbb2ï4ac

dAdx=2x+4(xy·+y)=0

y=2(2)ï(ïx)x2ï4(2)(ïx2)

2x+4[x(²2yx)+y]=0

y=4xx2+8x2

2x²8y+4y=0

y=4x3x

2x=4y

y=x and ï21x

Y Y Y

c
 
 c       


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

Problem 31

y = -½ x is absurd, thus use y = x

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

Therefore

y=aï4(61a)

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

P=2x+x=3x

y=31a

x=y=31P

Vmax=(31a)2(31a)

h=y2ï025x2

Vmax=1108a3 in3

answer

h=(31P)2ï41(31P)2 Solution:

h=91P2ï136P2

Problem 30

h=112P2

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

h=123P

A=dh dddA=ddddh+h=0

A=21xh

ddddh+h=0

Amax=21(31P)(123P)

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

Amax=1123P2 Amax=112333P2 [math]Amax=363P2

Convex surface area (shaded area):

dïdddh+h=0 answer

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

h=hd2 Solution:

h2=d2

From the figure:

d=h

D2=d2+h2

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.

0=2d+2hdddh

dddh=ïhd Volume of cylinder: Solution:

V=41d2h

a=4x+y

dhdV=4d2dddh+2dh=0

y=a²4x

ddddh+2h=0

Volume

dïhd+2h=0

V=x2y

2h=hd2

V=x2(a²4x)

d2=2h2

V=ax2²4x3

d=2h

dVdx=2ax²12x2=0

diameter =2 height

Solution: Area: A=xy From the figure: yaïx=ab answer y=ab(aïx) 2x(a²6x)=0

Y Y Y

c
 
 c       


c A=xab(aïx)

x=24 feet

dxdA=aba2+b2abï2xa2+b2=0

A=bxïabx2

y=100²1225(24)

2xa2+b2=ab

dxdA=bïa2bx=0

y=50 feet

x=ab2a2+b2

a2bx=b

dimensions: 50 ft × 24 ft

x=21a

Problem 34

y=aba2+b2abïab2a2+b2a2+b2

y=ab(aï21a)

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

y=aba2+b2abï2ab

y=21b dimensions: ½ a

answer

y=aba2+b22ab ×

½b

answer

y=2a2+b2

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.

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: Area: A=xy By similar triangle: cot=rx=ab r=abx Solution: Area: cot=sx=ba

Solution:

A=xy s=bax

Print Area:

By similar triangle:

(x²2)(y²3)=24

cot=rx=8060 r+y+s=a2+b2

y=24xï2+3

r=43x abx+y+bax=a2+b2 cot=sx=6080 y+aba2+b2x=a2+b2

Page area:

s=34x y=a2+b2ïaba2+b2x

A=xy

r+y+s=100 y=ababa2+b2ï(a2+b2)x

A=x24xï2+3

43x+y+34x=100 y=aba2+b2abïxa2+b2 y=100ï1225x

A=24xxï2+3x

Thus,

dxdA=(xï2)2(xï2)24ï24x(1)+3=0 ï48(xï2)2+3=0

A=x(100ï1225x) Thus,

²48+3(x²2)2=0

A=100xï1225x2 A=xaba2+b2abïxa2+b2

x=483+2

dxdA=100ï1250x A=aba2+b2abxïx2a2+b2

x=6 in

x=100(5012

Y Y Y

c
 
 c       


c Volume: dbdA=82(ï4)b+bdbdh+h=0

V=r2h

y=246ï2+3

41bïbï41b+h=0

0=r2drdh+2rh

y=9 in

h=b

drdh=ïr22rh

breadth = height 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?

answer

drdh=ïr2h

Problem 37

Let

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

m = price per unit area of glass

Solution:

k = total material cost per jar

½ m = price per unit area of plastic

From Solution of Problem 36 dbdh=ï41

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

r=21b

k=2mrh+23mr2 drdk=2mrdrdh+h+3mr=0

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

2rdrdh+h+3r=0 2rïr2h+h+3r=0

A=41r2+b(h²r) Solution: Given perimeter: P=b+2(h²r)+r P=b+2h²2r+r Where: b = 2r[/math] thus, r = ½ b

2h=3r A=41(21b)2+b(h²21b) h=23r A=116b2+bh²21b2 A=116(ï8)b2+bh

height = 3/2 × radius of base answer

dbdA=216(ï8)b+bdbdh+h=0

Problem 39

81bïbï41b+h=0

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?

ï(1+81)b+h=0 h=(1+81)b height=(1+81)breadth

P=b+2h²b+21b P=2h+21b 0=2dbdh+21 dbdh=ï41

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.

Light is most if area is maximum: A=21r2+b(h²r) A=21(21b)2+b(h²21b) A=81b2+bh²21b2 A=81(ï4)b2+bh

Solution:

Y Y Y

c
 
 c       


c 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+2bïa2=a2 Solution: sin45=h214ïa

Solution:

h=a2ï4(bïa)2

h=214ïasin45

h2+2Tïa2=2Lïa2

h=214ïa12

h=214a2ï(bïa)2

h=4(Lïa)2ï4(Tïa)2

Capacity is maximum if area is maximum:

h=21(Lïa)2ï(Tïa)2

A=21(b+a)h

h=21L2ï2La+a2ïT2+2Taïa2

A=21(b+a)214a2ï(bïa)2

h=21L2ïT2ï2(LïT)a

A=41(b+a)4a2ï(bïa)2 that a is constant)

(take note

h=2214ïa Area: A=A1+2A2 A=ah+2(21h2)

Area:

A=a2214ïa+2214ïa2

A=21(a+T)h

A=7a2ïa222+8(14ïa)2

A=21(a+T)21L2ïT2ï2(LïT)a dbdA=41(b+a)ï2(bïa)24a2ï(bïa)2+4a2ï (bïa)2=0

4a2ï(bïa)2=b2ïa24a2ï(bïa)2

A=41(a+T)L2ïT2ï2(LïT)a L and T are constant)

(note that

72ïa2ï27+4a=0

dadA=41(a+T)ï2(LïT)2L2ïT2ï2(LïT)a+L2ïT 2ï2(LïT)a=0 L2ïT2ï2(LïT)a=(a+T)(LïT)L2ïT2ï2(LïT)a

4a2ï(bïa)2=b2ïa2 4a2²b2+2ab²a2=b2²a2 2b2²2ab²4a2=0

L2²T2²2(L²T)a=(a+T)(L²T) L2²T2²2La+2Ta=La²Ta+TL²T2

3(L²T)a=L(L²T)

(b+a)(b²2a)=0

41ï12a=27ï72 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?

L2²TL=3La²3Ta

b2²ab²2a2=0

dadA=72ïa2ï414ïa=0

a=31L base = 1/3 × length of strip

answer

Problem 42 For b + a = 0; b = -a (meaningless) For b - 2a = 0; b = 2a (ok)

use b = 2a

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.

answer

Problem 41

Y Y Y

c
 
 c       


c Solution:

Solution:

D1=x2+62=x2+36

L1=x2+82=x2+64

D2=(10ïx)2+182=(10ïx)2+324

L2=(15ïx)2+122=(15ïx)2+144

Total Distance: D=D1+D2

Total length of wire:

D=x2+36+(10ïx)2+324

L=L1+L2

dxdD=2x2x2+36+2(10ïx)(ï1)2(10ïx)2+324 =0

L=x2+64+(15ïx)2+144

Solution: S1=x2+a2 S2=(cïx)2+b2

xx2+36=10ïx(10ïx)2+324 x(10ïx)2+324=(10ïx)x2+36

dxdL=2x2x2+64+2(15ïx)(ï1)2(15ïx)2+144= 0

Total distance traveled by light: S=S1+S2

x2[(10²x)2+324]=(10²x)2(x2+36) x2(10²x)2+324x2=x2(10²x)2+36(10²x)2

xx2+64=15ïx(15ïx)2+144

324x2=36(100²20x+x2)

dxdS=2x2x2+a2+2(cïx)(ï1)2(cïx)2+b2

9x2=100²20x+x2

x(15ïx)2+144=(15ïx)x2+64

8x2+20x²100=0

x2[(15²x)2+144]=(15²x)2(x2+64)

2x2+5x²25=0

x2(15²x)2+144x2=x2(15²x)2+64(15²x)2

(2x²5)(x+5)=0

144x2=64(225²30x+x2)

xx2+a2=cïx(cïx)2+b2 x(cïx)2+b2=(cïx)x2+a2 x2[(c²x)2+b2]=(c²x)2(x2+a2) x2(c²x)2+b2x2=x2(c²x)2+a2(c²x)2

9x2=4(225²30x+x2)

b2x2=a2(c2²2cx+x2)

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

5x2+120x²900=0

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

x2+24x²180=0

Use x = 5/2 = 2.5 mi

(x+30)(x²6)=0

D=252+36+(10ï25)2+324

For x + 30 = 0; x = ²30 (meaningless)

D=26 mi

For x - 6 = 0; x = 6 (ok)

answer

S=x2+a2+(cïx)2+b2

b2x2=a2c2²2a2cx+a2x2 (a2²b2)x2²2a2cx+a2c2=0 By Quadratic Formula: A = a2 ² b2; B = ²2a2c; C = a2c2 x=2AïBB2ï4AC

Problem 44

use x = 6 ft

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?

location of stake: 6 ft from the shorter post answer

x=2(a2²b2)2a2c4a4c2ï4(a2²b2)(a2c2) x=2(a2²b2)2a2c4a4c2ï4a4c2+4a2b2c2 x=2(a2²b2)2a2c2abc

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 ǂ = ǃ.

x=ac(ab)(aïb)(a+b) For x=ac(a+b)(aïb)(a+b) x=acaïb

meaningless if a > b

For x=ac(aïb)(aïb)(a+b) x=aca+b

ok

Y Y Y

c
 
 c       


c d=(xï0)2+(yïk)2 Use x=aca+b

d=x2+(yïk)2

by trial and error: x=a

when S is minimum:

d=a2+y2+(yïk)2

y=a2a3=a

cïx=cïaca+b

dydd=2y+2(yïk)2a2+y2+(yïk)2=0

cïx=a+b(a+b)cïac

2y+2(yïk)

cïx=a+bac+bcïac

2y=k

cïx=bca+b

y=21k

tan=xa

the nearest point is (a, a)

answer

y2=41k2

tan=aaca+b

Nearest Distance:

tan=ca+b

d=a2+41k2+(21kïk)2

tan=bcïx

d=a2+21k2

answer

tan=bbca+b tan=ca+b

Problem 47

tan ǂ = tan ǃ thus, ǂ = ǃ

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

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

Solution: d=(xï4a)2+(yï0)2 d=(xï4a)2+y2 from Solution: Standard equation: x2a2ïb2y2=1 For equilateral hyperbola, b = a. x2a2ïa2y2=1 x2²y2=a2

a2y=x3 y=x3a2 y2=x6a4 d=(xï4a)2+x6a4 dxdd=2(xï4a)+a46x5/2(xï4a)2+x6a4=0 (xï4a)+a43x5=0

x2=a2+y2 3x5+a4x²4a5=0 Distance d:

Y
Y Y