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Maxima and Minima Values by Second Derivation Method.
• Maxima and Minima of Two Functions of Two Indipendent Variables.
• In calculus, the second derivative test is a criterion for determining whether a given critical point of a real
function of one variable is a local maxmimum or a local minimum using the value of the second derivative at
the point.
The test states: if the function f is twice differentiable at a critical point x (i.e. f'(x) = 0), then:
If then f has a local maximum at 𝑥
If then f has a local minimum at 𝑥
If then the test is inconclusive.
• Absolute Mzxima and Minima.
The Absolute Maxima and Minima Values of z = f (x,y) indicate the over all largest and smallr=est values of f
in function’s domain.
• Stationary Points.
The points (x,y) for whitch df/dx = 0 and df/dy = 0 are called Stationary Points of f (x , y).
• All Points of Local Extrime anr Stationary Points.
• But, the Converse may or maynot be true.
• Critical Points.
Critical Points of f ( x, y) are Those Points Which are Either a Stationary Points or Where or both of fx or fy do not Exist.
• Like in the Case of Stationary Points , the Points Where one or both of fx or fy do not Exist , may Have an Extreme
Valyue.
• Saddle Point or Minimax Point.
• In mathematics, a saddle point is a point in the range of a function that is a stationary point but not a local extremum.
The name derives from the fact that the prototypical example in two dimensions is a surface that curves up in one
direction, and curves down in a different direction, resembling a saddle or a mountain pass. In terms of contour lines,
a saddle point in two dimensions gives rise to a contour that appears to intersect itself.
• Working Rules for Finding Extemum Values of a Function f ( x, y).
(a). Find df/dx and df/dy.
(b). Find Points ( x, y) for Whitchdf/dx = 0 and df/dy = 0. This Gives the Statiopnary Points.
(c). r =
𝑑2𝑓
𝑑𝑥2 , 𝑠 =
𝑑2𝑓
𝑑𝑥𝑑𝑦
, 𝑡 =
𝑑2𝑓
𝑑𝑦2
The Nature of Stationary Points are rt-𝑠2.
• Multivariable Case.
• For a function of more than one variable, the second derivative test generalizes to a test based on
the eigenvalues of the function's Hessian matrix at the critical point. In particular, assuming that all second order
partial derivatives of f are continuous on a neighbourhood of a critical point x, then if the eigenvalues of the
Hessian at x are all positive, then x is a local minimum. If the eigenvalues are all negative, then x is a local
maximum, and if some are positive and some negative, then the point is a saddle point. If the Hessian matrix
is singular, then the second derivative test is inconclusive.
• Concavity Test.
• A related but distinct use of second derivatives is to determine whether a function is concave up or concave
down at a point. It does not, however, provide information aboutinflection points. Specifically, a twice-
differentiable function f is concave up if and concave down if Note that if ,then
has zero second derivative, yet is not an inflection point, so the second derivative alone does not give enough
information to determine if a given point is an inflection point.
Examples
• Ex.1 Discribe The Nature of The Stationary Points For
f(x,y)=𝒙 𝟒
+ 𝒚 𝟒
− 𝟐𝒙 𝟐
+ 𝟒𝒙𝒚 − 𝟐𝒚 𝟐
df/dx = 4𝑥3
− 4𝑥 + 4𝑦 at ( -2 , 2 )
df/dy = 4𝑦3
+ 4𝑥 − 4𝑦
For Stationary points We Have df/dx = 0 ,df/dy = 0 , so…
𝑥3
− 𝑥 + 𝑦 = 0
𝑦3
+ 𝑥 − 𝑦 = 0
𝑥3
+ 𝑦3
= 0 𝑥 + 𝑦 𝑥2
− 𝑥𝑦 + 𝑦2
= 0
𝑥 + 𝑦 = 0 𝑦 = −𝑥
Putting 𝑦 = −𝑥
𝑥3
− 2𝑥 = 0 𝑥 𝑥2
− 2 = 0
(√2 , -√2) , (-√2 , √2) , ( 0,0) are Stationary Points
r = 12𝑥2
− 4 ,s = 4 ,t = 12𝑦2
− 4 ,t = 12𝑦2
− 4
rt-𝑠2
= 144𝑥2𝑦2
− 48𝑥2
− 48𝑦2
At (√2 ,-√2)
rt-𝑠2
= 384 > 0
r = 20>0
(√2 , √2) is a Minimum Point
rt-𝑠2
= 384 > 0
r = 12𝑥2
− 4 = 20 > 20
so ( -2 ,2) is inimum Point
at (0 , 0) rt-𝑠2
= 0
At (√2 ,-√2)
rt-𝑠2
= 384 > 0
r = 20>0
(√2 , √2) is a Minimum Point
rt-𝑠2
= 384 > 0
r = 12𝑥2
− 4 = 20 > 20
so ( -2 ,2) is inimum Point
at (0 , 0) rt-𝑠2
= 0
• Ex.2 Find The Extrime Value of sinx siny sin (x+y)
• df/dx = cosx siny sin (x+y) +sinx siny cos (x+y)
• =siny sin (2x+y)
• df/dy = sinx cosy sin (x+y +sinx siny cos (x+y)
• =sinx sin (x+2y)
• For Stationary Points we Have
• Df/dx = 0 ,df/dy = 0
• =siny [cosx (sin (x+y)]+sinx [cos (x+y )] = 0
• Tan (x+y) = -tanx ; x≠π/2 ,3π/2 ,x+y≠π/2 ,3π/2
• We have tanx = tany ;x = y
• Tan2x = -tanx = sin2x/cos2x = -sinx/cosx
• sin3x/cos2x cosx = 0
• sin3x = 0
• 3x = π
• r =
𝑑2 𝑓
𝑑𝑥2 = 2𝑦 cos
(2𝑥 + 𝑦)
• S =
𝑑2𝑓
𝑑𝑥𝑑𝑦
= 𝑐𝑜𝑠𝑥 sin 𝑥 + 2𝑦 + 𝑥 cos
(𝑥 + 2𝑦)
• T =
𝑓
𝑑𝑦2 = 2𝑠𝑖𝑛𝑥cos (x+2y)
• Thus, f(x,y) is ,aximum at (π/3 , π/3) and the Maximum Value is
• F (π/3 , π/3) = sin π/3 sin π/3 sin 2π/3
• = (√3/2) (√3/2) (√3/2)
• = 3√3/8
THANK YOU

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Calculus

  • 1. Maxima and Minima Values by Second Derivation Method. • Maxima and Minima of Two Functions of Two Indipendent Variables. • In calculus, the second derivative test is a criterion for determining whether a given critical point of a real function of one variable is a local maxmimum or a local minimum using the value of the second derivative at the point. The test states: if the function f is twice differentiable at a critical point x (i.e. f'(x) = 0), then: If then f has a local maximum at 𝑥 If then f has a local minimum at 𝑥 If then the test is inconclusive.
  • 2. • Absolute Mzxima and Minima. The Absolute Maxima and Minima Values of z = f (x,y) indicate the over all largest and smallr=est values of f in function’s domain. • Stationary Points. The points (x,y) for whitch df/dx = 0 and df/dy = 0 are called Stationary Points of f (x , y). • All Points of Local Extrime anr Stationary Points. • But, the Converse may or maynot be true. • Critical Points. Critical Points of f ( x, y) are Those Points Which are Either a Stationary Points or Where or both of fx or fy do not Exist. • Like in the Case of Stationary Points , the Points Where one or both of fx or fy do not Exist , may Have an Extreme Valyue. • Saddle Point or Minimax Point. • In mathematics, a saddle point is a point in the range of a function that is a stationary point but not a local extremum. The name derives from the fact that the prototypical example in two dimensions is a surface that curves up in one direction, and curves down in a different direction, resembling a saddle or a mountain pass. In terms of contour lines, a saddle point in two dimensions gives rise to a contour that appears to intersect itself.
  • 3. • Working Rules for Finding Extemum Values of a Function f ( x, y). (a). Find df/dx and df/dy. (b). Find Points ( x, y) for Whitchdf/dx = 0 and df/dy = 0. This Gives the Statiopnary Points. (c). r = 𝑑2𝑓 𝑑𝑥2 , 𝑠 = 𝑑2𝑓 𝑑𝑥𝑑𝑦 , 𝑡 = 𝑑2𝑓 𝑑𝑦2 The Nature of Stationary Points are rt-𝑠2.
  • 4. • Multivariable Case. • For a function of more than one variable, the second derivative test generalizes to a test based on the eigenvalues of the function's Hessian matrix at the critical point. In particular, assuming that all second order partial derivatives of f are continuous on a neighbourhood of a critical point x, then if the eigenvalues of the Hessian at x are all positive, then x is a local minimum. If the eigenvalues are all negative, then x is a local maximum, and if some are positive and some negative, then the point is a saddle point. If the Hessian matrix is singular, then the second derivative test is inconclusive. • Concavity Test. • A related but distinct use of second derivatives is to determine whether a function is concave up or concave down at a point. It does not, however, provide information aboutinflection points. Specifically, a twice- differentiable function f is concave up if and concave down if Note that if ,then has zero second derivative, yet is not an inflection point, so the second derivative alone does not give enough information to determine if a given point is an inflection point.
  • 5. Examples • Ex.1 Discribe The Nature of The Stationary Points For f(x,y)=𝒙 𝟒 + 𝒚 𝟒 − 𝟐𝒙 𝟐 + 𝟒𝒙𝒚 − 𝟐𝒚 𝟐 df/dx = 4𝑥3 − 4𝑥 + 4𝑦 at ( -2 , 2 ) df/dy = 4𝑦3 + 4𝑥 − 4𝑦 For Stationary points We Have df/dx = 0 ,df/dy = 0 , so… 𝑥3 − 𝑥 + 𝑦 = 0 𝑦3 + 𝑥 − 𝑦 = 0 𝑥3 + 𝑦3 = 0 𝑥 + 𝑦 𝑥2 − 𝑥𝑦 + 𝑦2 = 0 𝑥 + 𝑦 = 0 𝑦 = −𝑥 Putting 𝑦 = −𝑥 𝑥3 − 2𝑥 = 0 𝑥 𝑥2 − 2 = 0 (√2 , -√2) , (-√2 , √2) , ( 0,0) are Stationary Points r = 12𝑥2 − 4 ,s = 4 ,t = 12𝑦2 − 4 ,t = 12𝑦2 − 4 rt-𝑠2 = 144𝑥2𝑦2 − 48𝑥2 − 48𝑦2 At (√2 ,-√2) rt-𝑠2 = 384 > 0 r = 20>0 (√2 , √2) is a Minimum Point rt-𝑠2 = 384 > 0 r = 12𝑥2 − 4 = 20 > 20 so ( -2 ,2) is inimum Point at (0 , 0) rt-𝑠2 = 0 At (√2 ,-√2) rt-𝑠2 = 384 > 0 r = 20>0 (√2 , √2) is a Minimum Point rt-𝑠2 = 384 > 0 r = 12𝑥2 − 4 = 20 > 20 so ( -2 ,2) is inimum Point at (0 , 0) rt-𝑠2 = 0
  • 6. • Ex.2 Find The Extrime Value of sinx siny sin (x+y) • df/dx = cosx siny sin (x+y) +sinx siny cos (x+y) • =siny sin (2x+y) • df/dy = sinx cosy sin (x+y +sinx siny cos (x+y) • =sinx sin (x+2y) • For Stationary Points we Have • Df/dx = 0 ,df/dy = 0 • =siny [cosx (sin (x+y)]+sinx [cos (x+y )] = 0 • Tan (x+y) = -tanx ; x≠π/2 ,3π/2 ,x+y≠π/2 ,3π/2 • We have tanx = tany ;x = y • Tan2x = -tanx = sin2x/cos2x = -sinx/cosx • sin3x/cos2x cosx = 0 • sin3x = 0 • 3x = π • r = 𝑑2 𝑓 𝑑𝑥2 = 2𝑦 cos (2𝑥 + 𝑦) • S = 𝑑2𝑓 𝑑𝑥𝑑𝑦 = 𝑐𝑜𝑠𝑥 sin 𝑥 + 2𝑦 + 𝑥 cos (𝑥 + 2𝑦) • T = 𝑓 𝑑𝑦2 = 2𝑠𝑖𝑛𝑥cos (x+2y) • Thus, f(x,y) is ,aximum at (π/3 , π/3) and the Maximum Value is • F (π/3 , π/3) = sin π/3 sin π/3 sin 2π/3 • = (√3/2) (√3/2) (√3/2) • = 3√3/8