Application Problems in Engineering

APPLICATION PROBLEMS IN ENGINEERING

 

Example 54. Find the minimum value of x²+ y²+z² when ax + by + cz = p.

Solution: Let f = x² + y² + z²

 Ø = ax + by + cz ‒ p

 u = f +λØ = (x² + y²+z²) + λ(ax+by+cz‒p)

∂u/∂x = 2x +λa = 0  ... (1)

∂u/∂y = 2y+λb=0    ... (2)

∂u/∂z = 2z + λc = 0   …..(3)

∂u/∂λ = ax + by + cz‒p=0     ...(4)

From (1) we get x = ‒λa / 2        ...5)

From (2) we get y = ‒ λb / 2       ... (6)

From (3) we get z = ‒λc / 2         ... (7)

Substitute the values of x, y, z in (4) we get, a(‒λa / 2) + b(‒λb / 2) + c (‒λc / 2) = p;

 f = p² / (a² + b² + c²)

 

Example 55. Find the minimum value of x² + y²+z² when x+y+z=3a.

Solution:

Let f = x² + y² + z²

 ϕ=x+y+z‒3a

 u = f +λϕ = x² + y² + z² + λ(x + y + z‒3a)

∂u/∂x = 2x+λ = 0  .. (1)

∂u/∂z = 2y+ λ =0   ......(2)

∂u/∂y = 2z+λ=0   …(3)

∂u/∂λ = x+y+z‒3a = 0         ...... (4)

From (1) we get x = ‒λ/2         ...... (5)

From (2) we get y = ‒λ/2         ……..(6)

From (3) we get z = ‒λ/2       ……...(7)

Substitute the values of x, y, z in (4) we get, (‒λ/2) + (‒λ/2) + (‒λ/2) = 3a

 (‒3λ / 2) = 3a

 ∴ λ = ‒2a

Substitute the values of λ in (5), (6) and (7) we get,

x = a; y = a; z = a

 ∴ The minimum point is (a, a, a)

The minimum value of f = x² + y² + z² = a² + a² + a²

 f = 3a²

 

Example 56. If u = (a³x²+b³y²+c³z²), where x^‒1+y^‒1+z^‒1=1, show that the stationary values of u is given by x=Σa/a; y=Σb/b; z=Σc/c;

Solution:

Let f = a³x² + b³y² + c³z²

 ϕ = 1/x + 1/y + 1/z ‒ 1

 u = f + λϕ = (a³x² + b³y² + c³z²) + λ(1/x + 1/y + 1/z ‒ 1)

The stationary values are x=Σa/a; y=Σb/b; z=Σc/c;

 

Example 57. Find the maximum value of x^myⁿz^P when x + y + z = a.

Solution:

Let f = x^myⁿz^p

 ϕ = x+y+z‒a

 u = f + λϕ = x^myⁿz^P + λ (x + y + z‒a)

NOTE: If the Auxillary function u = f + λϕ contains the term xyz, then to find the values of x, y, z, we use the technique ∂u/∂x × x ‒ ∂u/∂y × y = 0 and ∂u/∂x × x ‒ ∂u/∂z × z = 0.

 

Example 58. Find the minimum value of x² + y²+z² when xyz = a³.

Solution:

Let f = x² + y² + z²

 ϕ = xyz ‒ a³

 u = f + λϕ = x² + y² + z² + λ(xyz‒a³)

∂u/∂x = 2x + λyz = 0           ..... (1)

∂u/∂y = 2y + λxz = 0            .... (2)

∂u/∂z = 2z + λxy = 0              …(3)

∂u/∂λ = xyz ‒ a³ = 0                ...(4)

(1)×x − (2)×y ⇒ 2x² + λxyz ‒ 2y² ‒ λxyz = 0

2x² = 2y²   ⇒  y= x              ………..(5)

(1)×x − (3)×z ⇒ 2x² + λxyz ‒ 2z² – λxyz = 0

2x² = 2z² ⇒ z = x      .....(6)

Substitute y, z in (4) we get, x³ = a³ ⇒ x= a

(5) ⇒ y = a;

(6) ⇒ z = a

Minimum point is (a, a, a)

The minimum value f = x²+ y²+z² = a² + a² + a²

The minimum value f = 3a²

 

Example 59. A rectangular box, open at the top is to have the volume of 32c.c. Find the dimensions of the box that requires the least material for its construction.

Solution:

 Let the dimensions of the box be x, y, z

Given volume = 32cc i.e. xyz = 32

Let ϕ = xyz‒32

Least material for its construction means to minimize the surface area.

Surface area = xy + 2yz + 2zx

Let f = xy +2yz + 2zx

 u = f + λø = (xy+2yz +2zx) + λ(xyz‒32)

∂u/∂x = y + 2z+λyz = 0            ...(1)

∂u/∂y = x + 2z + λxz = 0              ...(2)

∂u/∂z = 2y + 2x + λxy = 0              ...(3)

∂u/∂λ = xyz‒32=0                     ...(4)

 (1)×x ‒ (2)×y = xy + 2xz + λxyz ‒ xy‒2yz ‒ λxyz = 0

2xz = 2yz ⇒ y = x       .....(5)

 (1)×x ‒ (3)×z = xy + 2xz + λxyz ‒ 2yz ‒ 2xz ‒ λxyz = 0

 xy = +2yz ⇒ x= 2z

i.e. z = x/2               .....(6)

 (4) = x (x) (x/2) = 32

 x³ = 64      ⇒      x = 4

(5) ⇒ y = 4;

(6) ⇒ z = 2

∴ Length = 4cm;

Breadth = 4cm;

Height = 2cm

 

Example 60. Find the dimensions of a rectangular box, without top of maximum capacity and surface area 432 square meters.

Solution: Let the dimensions of the box be x, y, z

Given surface area = 432

xy+2yz + 2zx = 432

Let ϕ = xy + 2yz + 2zx ‒ 432

We have to maximize the capacity (Volume)

Volume = xyz.

Let f = xyz

 u = f + λϕ = xyz + λ(xy + 2yz + 2zx − 432)

∂u/∂x = yz + λy + 2λz = 0        ... (1)

∂u/∂y = xz + λx + 2λz = 0      ... (2)

∂u/∂z = xy + 2λy + 2λx = 0       ...(3)

∂u/∂λ = xy + 2yz + 2zx ‒ 432 = 0... (4)

 (1)×x ‒ (2)×y

⇒ xyz + λxy+2λxz‒xyz ‒ λxy ‒ 2λyz = 0

 2λxz = 2λyz

⇒ y = x                 ...... (5)

(1)×x − (3)×z

⇒ xyz + λxy + 2λxz ‒ xyz ‒ 2λyz – 2λxz = 0

 λxy = 2λzy

z = x/2           ….(6)

Substitute the values of y, z in (4) we get,

 x² + 2x(x/2) + 2(x/2)x ‒ 432= 0

 3x² = 432 ⇒ x² = 144

x = 12

(5) ⇒ y = 12

(6) ⇒ z = 12/2 = 6

Length = 12m;

breadth = 12m;

height = 6m

 

Example 61. Find the volume of the greatest rectangular parallelepiped inscribed in the ellipsoid whose equation is x²/a² + y²/b² + z²/c² =1

Solution: Let 2x, 2y, 2z be the dimensions of the rectangular box.

Maximum point is = 8abc / 3√3

 

Example 62. Prove that the rectangular solid of maximum volume which can be inscribes in a sphere is a cube.

Solution:

Let the equation of the sphere be x² + y² + z² = a²

Let ϕ = x² + y² + z² ‒ a²

Let 2x, 2y, 2z be the dimensions of the rectangular parallelopiped.

Volume 2x 2y 2z = 8xyz

Let f = 8xyz

 u =  f + λϕ = 8xyz + λ(x² + y² + z² ‒ a²)

∂u/∂x = 8yż + λ2x = 0    ..... (1)

∂u/∂y = 8xz + λ2y = 0     ..... (2)

∂u/∂z = 8xy + λ2z = 0     ...(3)

∂u/∂λ = x² + y² + z² ‒ a² = 0    ... (4)

 (1)×x − (2)×y ⇒ y=x         ……….(5)

 (1)×x − (3)×z ⇒ z = x        ……..(6)

Substitute the values of y and z in (4), we get

 x² + x² + x² ‒ a² = 0

x² = a²/3

i.e. x = a/√3

 (5) ⇒ y = a/√3

(6) ⇒ z = a/√3

∴ x = y = z = a/√3

Since the three sides are equal, the given surface is a cube.

 

Example 63. The temperature T(x, y, z)at any point in space is T = 400 xyz². Find the highest temperature on the surface of the sphere x² + y² + z² = 1.

Solution: Let f = 400xyz²; ϕ= x² + y²+z²‒1

 u = f + λϕ = (400 x y z²) + λ(x² + y² + z² − 1)

∂u/∂x = 400yz² + 2λx = 0       ...(1)

∂u/∂y = 400xz² + 2λy = 0     ... (2)

∂u/∂z = 800xyz +2λz = 0        ... (3)

∂u/∂λ = x² + y² + z² − 1 = 0       ... (4)

(1)×x ‒ (2)×y ⇒ 400xyz² + 2λx² ‒ 400xyz² ‒ 2λy²=0

x² = y² ⇒ y = x     ……(5)

 (1)×2x − (3)×z ⇒ 800xyz² + 4λx² ‒ 800xyz² ‒ 2λz² = 0

4λx² = 2λz²  ⇒  z = √2x       ... (6)

Substitute the values of y, z in (4) we get, x² + x² + 2x² = 1

x² = 1/4,

x = 1/2,

(5) ⇒ y = 1/2

(6) ⇒ z = 1/√2

The highest temperature is f = 400xyz²

 f = 400(1/2)(1/2)(1/2) = 50

The highest temperature is 50°C

 

Example 64. Find the greatest and the least distances of the point (3,4,12) from the unit sphere whose center is at the origin.

Solution: Equation of unit sphere with center at the origin is x² + y²+z² = 1

Let Q(3,4,12) be the given point and P(x, y, z) be any point on the unit sphere

 d = PQ = √[ (x−3)² + (y−4)² + (z–12)² ]

We have to minimize d i.e. minimize d²

 f = (x − 3)² + (y−4)² + (z ‒ 12)²;

 ϕ = x² + y² + z² ‒ 1

 u = f + λϕ = [(x‒3)² + (y‒4)² + (z‒2)²] + λ(x² + y²+z²‒1)

∂u/∂x = 2(x‒3) + 2λx = 0          ... (1)

∂u/∂y = 2(y‒4) + 2λy = 0         ... (2)

∂u/∂z = 2(z‒12) + 2λz = 0        ... (3)

∂u/∂λ = x² + y² + z² − 1 = 0       ... (4)

 (1) ⇒ x = 3/(1+λ)         ……(5)

 (2) ⇒ y = 4/(1+λ)         ……(6)

 (3) ⇒ z = 12/(1+λ)         ……(7)

Substitute x,y,z in (4) we get,

Shortest distance = 12; Longest distance = 14

 

Example 65. Find the length of the shortest distance from the point (0,0,25/9) to the surface z = xy.

Solution: Given surface is z = xy.

Let Q(0,0,25/9) be the given point and P(x, y, z) be any point on the given surface.

 d = PQ = √[ (x − 0)² + (y − 0)² + (z ‒ 25/9)² ]

We have to minimize d i.e. minimize d²

 f = x² + y² + (z – 25/9)²;

 ϕ = z − xy

 u = f + λϕ = x² + y² + (z – 25/9)² + λ(z − xy)

∂u/∂x = 2x‒λy= 0   ... (1)

∂u/∂y = 2y‒λx = 0    ... (2)

∂u/∂z = 2(z ‒ 25/9) + λ = 0     ... (3)

∂u/∂λ = z − xy = 0     ... (4)

 (1) ⇒ λ = 2x/y        ... (5)

 (2) ⇒ λ = 2y/x        ... (6)

From (5) and (6), we get

2x / y = 2y / x

⇒ x² = y²

 x = ±y               ... (7)

Case 1: x = y

 (5) ⇒ λ = 2

 (4) ⇒ z = x²          ... (8)

 (3) ⇒ 2(z ‒ 25/9) + 2 = 0

2(z ‒ 25/9) = ‒2

 z ‒ 25/9 = ‒1

z  = ‒1 + 25/9

z  = 16/9

 (8) ⇒  x² = 16/9

x = ± 4/3

y = ± 4/3

 d = √[ (4/3)² + (4/3)² + (16/9 ‒ 25/9)² ]

 = √41 / 3

Case 2x = ‒y

(5)  ⇒ λ = ‒2

(4) ⇒ z = ‒ x²           ... (9)

(3) ⇒ 2(z ‒ 25/9) ‒2 = 0

2(z‒ 25/9) = 2

 z ‒ 25/9 = 1

 z = 1 + 25/9

 z = 34/9

 (9) ⇒ x² = ‒34 / 9

which gives a complex value. Hence this case is not possible.

.. Shortest distance = √41 / 3

 

Example 66. Find the minimum distance of the point (1,2,0) from the cone x²+ y² ‒ z² = 0.

Solution: Let P(x, y, z) be the point on the cone therefore the distance between (x, y, z) and Q(1,2,0)

 d = PQ = √ [ (x‒1)² + (y‒2)² + (z‒0)² ]

Let u = (x‒1)² + (y‒2)² + (z)² subject to x² + y² ‒ z² = 0

We have to minimize d i.e. minimize d²

 f = (x‒1)² + (y‒2)² + (z)²;

 ϕ = x²+ y² ‒ z²

 u = f + λϕ = [(x‒1)² + (y‒2)² + (z)²]+ λ(x²+y²‒z²)

∂u/∂x = 2(x‒1) + 2λx = 0    ... (1)

∂u/∂y = 2(y − 2) + 2λy = 0    ... (2)

∂u/∂z = 2z ‒ 2λz = 0            ... (3)

∂u/∂λ = x²+ y² ‒ z² = 0...(4)

 (1) ⇒ x = 1/ (1+λ)          ……(5)

(2) ⇒ y = 2 / (1+λ)          ……(6)

(3) ⇒ 2z(1‒λ) = 0

 z≠0,λ=1.         ……(7)

Substitute x, y, z in (4) we get,

(5) ⇒ x = 1/2

(6) ⇒ y=1

substituting the values of x,y in

 x²+ y²‒z²

we get

z = ± √5/2

 d = √ [ ( 1/2 − 1)² + (1‒2)² + 5/4 ] = √(10/4) = √(5/2)

Minimum distance = 12

 

 

EXERCISE

 

62. Examine the extreme values of the following

 

63. Find the points on the surface z² = xy + 1 nearest to the origin.

Ans: (0,0,1), (0,0, ‒1),

 

64. Find the minimum value of the function x² + y² + z² subject to the condition:

(i) ax + by + cz = a+b+c Ans: ((a+b+c)²) / a²+b²c² )

(ii) xy + yz + zx = 3a² Ans: 3a²at (a, a, a) and (‒a, ‒a, ‒a)

 

65. Find the minimum value of the function ax + by + cz subject to the condition a²/x + b²/y + c²/z = 1 Ans: f = (a^3/2 + b^3/2 + c^3/2)²

 

66. Prove that of all the rectangular parallelopipeds of the same volume, the cubes have the least surface.

 

67. Prove that of all rectangular parallelopipeds of given surface cube has the maximum volume.

 

68. The temperature T at any point (x, y, z) in space is T = 400xyz². Find the highest temperature on the surface of the sphere x² + y² + z² = a² Ans: ka⁴ / 8

In a triangle ABC, find the maximum value of cos A cos B cos C. Ans: f = 1/8

 

69. Find the shortest distance from the point (1,2,2) to the sphere x² + y² + z² = 36 Ans: 3

 

70. Find the shortest and longest distances from the point (1,2,‒1) to the sphere x² + y² + z² = 24 Ans: √6, 3√6

 

71. Find the dimensions of the rectangular box without top of maximum capacity and surface area 108 square meters. Ans: l = 6cm, b = 6cm, h = 3cm

 

72. Find the dimensions of the rectangular parallelopiped without top of given surface area, which has greatest volume.

Ans: Length = √(k/3); breadth = √(k/3); height = ½√(k/3)

 

73. A rectangular box opens at the top, is to have a given capacity. Find the dimensions of the box requiring least material for construction. (or)

A rectangular box opens at the top, is to have a given volume. Find the dimensions so that the surface area of the tank is minimum.

Ans: Length = 2^1/3k^1/3; breadth = 2^1/3k^1/3; height = 2^‒2/3k^1/3;

 

74. Find the foot of the perpendicular from the origin on the plane 2x + 3y ‒ z = 5. Ans: The foot of the perpendicular = (5/7, 15/14, ‒5/14)

 

75. Find the extreme value of the function x² + y² + z² subject to the condition: Ans: x + y + z = 3a

 

76. Divide 24 into three parts such that the continued product of first, square of the second and cube of the third may be maximum. Ans: 4,8,12