Advanced Engineering Mathematics,Applications,problems of Maxima and Minima andLagranges Theorem

1. ADVANCED ENGINEERING
MATHEMATICS
Submitted by:
Name : M.SAI SASIDHAR
Regd No.: 221FA18091
Batch : 2
Sec. : 23(AIML)
Submitted to:
Dr.S.H.Manjula

2. MAXIMA AND MINIMA
Maxima and minima are the peaks and
valleys in the curve of a function. There can
be any number of maxima and
minima(local) for a function. In calculus, we
can find the maximum and minimum value
of any function without even looking at the
graph of the function. Maxima will be the
highest point on the curve within the given
range and minima would be the lowest
point on the curve.
sasidhar

3. MAXIMA AND MINIMA OF TWO VARIABLES
MAXIMA AND MINIMA OF TWO VARIABLES
• Let f(x,y) be a function of two independent variables x and y, which is
continuous for all values of x and y in the neighborhood of (a,b) i.e.
(a+h,b+k) be a point in its neighborhood which lies inside the region R.
• The point (a,b) is called a point of relative minimum, if f(a,b) < f(a+h,b+k)
for all h,k Then f(a,b) is called the relative minimum value.
• The point (a,b) is called a point of relative maximum, if f(a,b) >
f(a+h,b+k) for all h,k Then f(a,b) is called the relative maximum value.
sasidhar

4. MAXIMA AND MINIMA OF TWO VARIABLES
• Stationary point: The point at which function
is either maximum or minimum is known as stationary point.
• Extreme Value: The value of the function at stationary point is known as
extreme value of the function.
• Saddle point : A point on the surface of the graph of a function where
the slopes (derivatives) in orthogonal directions are all zero (a critical
point), but which is not a local extremum of the function.
sasidhar

5. MAXIMA AND MINIMA OF TWO VARIABLES
Step 3:
1) If ln − 𝑚2
> 0 & l < 0 at (a,b), then f(x, y) is maximum at (a,b) &
maximum value of the function is f (a,b).
2) If ln − 𝑚2>0 & l > 0 at (a,b), then f(x,y) is maximum at (a,b) &
minimum value of the function is f (a,b).
3) If ln − 𝑚2< 0 at (a,b), then f (x,y) is neither maximum nor minimum
at (a,b).Such point is called Saddle Point.
4)If ln − 𝑚2 < 0 at (a,b), then no conclusion can be made about the
extreme values of f(x, y) & further investigation is required.
sasidhar

6. APPLICATIONS OF MAXIMA AND MINIMA
APPLICATIONS OF MAXIMA AND MINIMA IN DAILY
LIFE
Such applications exist in economics, business, and
engineering. Many can be solved using the methods
of differential calculus described above. For example,
in any manufacturing business it is usually possible to
express profit as a function of the number of units
sold.
sasidhar

7. PROBLEM
Show that the function f(x , y) = x3 + y3 -63(x + y) +12xy is maximum at
(-7 , -7) and minimum at (3 , 3). And also find maximum and minimum
values ?
Solution:
Given,
f(x , y) = x3 + y3 -63(x + y) +12xy
Now from the given function
𝜕𝑓
𝜕𝑥
= 3x2 – 63 + 12y
And
𝜕𝑓
𝜕𝑦
= 3y2 – 63 + 12x
sasidhar

8. PROBLEM
sasidhar
Let ∂f/∂x = 0 (1) and ∂f/∂y = 0 (2)
3x2 – 63 + 12y = 0
3y2 – 63 + 12x = 0
We can also write it as,
3(x2 + 4y – 21) = 0 x2 + 4y – 21 = 0 (3)
3(y2 + 4x – 21) = 0 y2 + 4x – 21 = 0 (4)
By solving the above two equations, we get
x2 – y2 – 4(x – y) = 0
(x – y) (x + y – 4) = 0

9. PROBLEM
x = y (5) and x + y = 4 (6)
Now equation(5) in equation(3) then it becomes
x2 + 4y – 21 = 0
sasidhar
x2 + 7x – 3x – 21 = 0
x(x + 7) – 3(x + 7) = 0
(x – 3) (x + 7) = 0
x = 3 , -7
x values in equation (4) it becomes
if x = -7 then y = -7
if x = 3 then y = 3
The two stationary points are (-7,-7) and (3,3)

10. PROBLEM
We know that l =
𝜕2𝑓
𝜕𝑥2
= 6x
m =
𝜕2𝑓
𝜕𝑥𝜕𝑦
= 12
n =
𝜕2𝑓
𝜕𝑦2
= 6y
Now ln – m2
6x(6y) – (12)2
36xy – 144
At point (-7 , -7)
l = 6x = 6(-7) = -42
sasidhar
ln - m2
= 36xy – 144 = 36(-7)(-7) – 144 = 1764 – 144 = 1620
In this point l < 0 ; ln - m2 > 0
At point (3 , 3)
l = 6x = 6(3) = 18
ln - m2
= 36xy – 144 = 36(3)(3) – 144 = 324 – 144 = 180
In this point l > 0 ; ln - m2 > 0

11. PROBLEM
At point (-1 , 5)
l = 6x = 6(-1) = -6
ln - m2
= 36xy – 144 = 36(-1)(5) – 144 = -180 – 144 = -324
In this point l < 0 ; ln - m2 < 0
And At point (5 , -1)
l = 6x = 6(5) = 30
ln - m2
= 36xy – 144 = 36(5)(-1) – 144 = -180 – 144 = -324
In this point l > 0 ; ln - m2 < 0
f(x ,y) has maximum value at (-7 , -7) and minimum value at
(3 , 3)
Now to find the maximum and minimum value we
substitute the points in given function
f(x , y) = x3 + y3 -63(x + y) +12xy
f(-7 , -7) = (-7)3 +(-7)3 – 63(-7-7) + 12(-7)(-7)
= -343-343+882-588
= -392
The maximum value is -392
f(x , y) = x3 + y3 -63(x + y) +12xy
f(3 , 3) = (3)3 + (3)3 – 63(3 + 3) + 12(3)(3)
= 27 + 27 – 378 + 108
= -216
The minimum value is -216.
sasidhar

12. LAGRANGES THEOREM
OF UNKNOWN MULTIPLIERS TO FIND MAXIMA AND MINIMA
sasidhar
Lagrange’s Method :
We use Lagrange’s method to find Stationary Points of a function
having several independent Variables.
We use this method to find extreme values of function subject to
several condition
Drawback (or) Remark.
Lagrange’s method fails to find nature of the stationary given
function. Points of a given function.
Working rule.
Let the Given function be f (x, y, z), ∅ 𝑥, 𝑦, 𝑧 = 0 ----(1)
Step-1:
Form Lagrangean function.
F(x, y, z) = f(x, y, z) +λ∅(x, y, z) where is λ Lagrangean
multiplier.
Step-2:
𝜕𝑓
𝜕𝑥
=0, →
𝜕𝑓
𝜕𝑥
+ λ
𝜕∅
𝜕𝑥
=0 –(2)
𝜕𝑓
𝜕𝑦
=0 →
𝜕𝑓
𝜕𝑦
+ λ
𝜕∅
𝜕𝑦
=0—(3)
𝜕𝑓
𝜕𝑧
=0 →
𝜕𝑓
𝜕𝑧
+ λ
𝜕∅
𝜕𝑧
=0—(4)
Step-3
Solve 2,3&4 to get stationary points
Step-4
substitute x, y, z in –(1) remove λ
step-5.
To find extremum substitute stationary Points in given
function f(x, y, z)

13. PROBLEM
sasidhar
Suppose you are designing a cylindrical container to hold a fixed volume V of
liquid. You want to minimize the cost of the container subject to the constraint
that the volume of liquid it holds is fixed.
Solution:
The problem you are trying to solve is called an optimization problem.
The volume of a cylinder is given by the formula 𝑉 = 𝜋𝑟2ℎ, where 𝑟 is the radius of the cylinder and
ℎ is its height. Since we want to minimize the cost of the container subject to the constraint that it
holds a fixed volume of liquid, we can express the cost of the container in terms of 𝑟 and ℎ, and then
use the volume constraint to eliminate one of these variables.
Minimize: Cost of the container, which is proportional to its surface area
Subject to: The volume of the container must be fixed at V
Variables: 𝑟 and ℎ the radius and height of the cylinder

14. PROBLEM
sasidhar
Using the formula for the volume of a cylinder, we can eliminate one of
the
variables. For example, we can solve the formula for h in terms of r and
V :
𝑉 = 𝜋𝑟2ℎ
Use the method of Lagrange multipliers to find the values of 𝑟 and ℎ that
minimize the cost of the container subject to the constraint that its volume is V.
To apply the method of Lagrange multipliers, we need to define the Lagrangian
function 𝑟, ℎ andλλ as follows:
L(r,h,λ)=2πrh+2π𝑟2 + λ(V−πr2h)

15. PROBLEM
sasidhar
Using the formula for the volume of a cylinder, we can eliminate one of
the
variables. For example, we can solve the formula for h in terms of r and
V :
𝑉 = 𝜋𝑟2
ℎ
ℎ =
𝑉
𝜋𝑟2

16. PROBLEM
sasidhar
Using the formula for the volume of a cylinder, we can eliminate one of
the
variables. For example, we can solve the formula for h in terms of r and
V :
𝑉 = 𝜋𝑟2
ℎ
ℎ =
𝑉
𝜋𝑟2

17. PROBLEM
sasidhar
Using the formula for the volume of a cylinder, we can eliminate one of
the
variables. For example, we can solve the formula for h in terms of r and
V :
𝑉 = 𝜋𝑟2
ℎ
ℎ =
𝑉
𝜋𝑟2

18. PROBLEM
sasidhar
F= 400 xz2 + λ(2y) = 0 4
∂s∂z+ λ ∂F∂z =0
F = 800 xyz + λ(2z) = 0 5
Multiply equation 3 with x;
Multiply equation 4 with y;
Multiply equation 5 with z;
we have
1600 xyz2 + 2 λ (x2 + y2 + z2) = 0
1600 xyz2 + 2 λ (1) = 0 (Given)
λ = -800 xyz2
substituting the value of λ in (3),
we get
400 yz2 + 2x (-800 xyz2)= 0⇒1-4y2= 0
⇒ y=±
1
2
substituting the value of λ in (5),
we get
400 yz2 + 2x (-800 xyz2) = 0
⇒1-4z2= 0
⇒ z=±
1
√2
Substituting x, y and z in s,
We get
S=400*
1
2
*
1
2
*
1
2
=50.
Therefore the maximum temperature of
surface of sphere is 50

19. THANK YOU
sasidhar