Maxima and Minima of Functions of Several Variables
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Maxima and Minima of Functions of Several Variables
Maxima and Minima of Functions of Several Variables
For calculating maxima and minima of functions of several variables we use
following steps:
Step 1: Let f(x) as a function then first of all we calculate differentiation of
function with respect to one variable like we choose x as a variable, so we can
find its derivative as: f1(x) = D f(x) Dx
Step 2: Now, put f1(x) equal to 0: f1(x) = 0.
Step 3: Then calculate critical points from f1(x) = 0: Suppose we get x =a and
y= b as a critical points.
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Step 4: Then we calculate second order differentiation of function f(x) means
f11(x)  f11(x) = D f1(x) Dx
Step 5: After this, we calculate value of f11(x) for critical points [f11(x)](a,b) = ?
Step 6: if value of [f11(x)](a,b) is positive then critical points are local minima of
function and if value of [f11(x)](a,b) is negative then critical points are local
maxima of function .
We take an example to understand maxima and minima of a function with
several variables.
Problem: Find maxima and minima of function f(x1,x2) = x12 + 5x2 + 6
Solution: Step 1: f1(x1,x2) = D f(x1,x2) = 2.x1 + 5x2 Dx1
Step 2: f1(x1,x2) = D f(x1,x2) = 0 => 2.x1 + 5x2 = 0 Dx1 => x1 = 5x2 2
Step 3: Critical points of function f(x1,x2) = > if x2 = 1 then x1 =5/2 and if x2 =
2 then x1 =5. So, we assume we have critical points (1,5/2) and (2, 5)......etc
Step 4: Then we calculate second order derivative of function f(x)  f11(x1) = D
f1(x1) = 2 + 5 x2 Dx1
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Step 5: At last we calculate value of f11(x1) for critical points  [f11(x)](1,5/2) =
2 + 5x2 = 2 + 5(5/2) = 2  25/2 = 21/2 (1,5/2) produces negative value, so
these critical points are called as a local maxima.
[f11(x)](1,5/2) = 2 + 5x2 = 2 + 5(5) = 2 + 25 = 27 (5,2) produces positive value
so, these critical points are called as a local minima.
ThankYou
TutorCircle.com
Maxima and Minima of Functions of Several Variables
For calculating maxima and minima of functions of several variables we use
following steps:
Step 1: Let f(x) as a function then first of all we calculate differentiation of
function with respect to one variable like we choose x as a variable, so we can
find its derivative as: f1(x) = D f(x) Dx
Step 2: Now, put f1(x) equal to 0: f1(x) = 0.
Step 3: Then calculate critical points from f1(x) = 0: Suppose we get x =a and
y= b as a critical points.
Know More About Rational Numbers Additive Inverse Worksheets
Step 4: Then we calculate second order differentiation of function f(x) means
f11(x)  f11(x) = D f1(x) Dx
Step 5: After this, we calculate value of f11(x) for critical points [f11(x)](a,b) = ?
Step 6: if value of [f11(x)](a,b) is positive then critical points are local minima of
function and if value of [f11(x)](a,b) is negative then critical points are local
maxima of function .
We take an example to understand maxima and minima of a function with
several variables.
Problem: Find maxima and minima of function f(x1,x2) = x12 + 5x2 + 6
Solution: Step 1: f1(x1,x2) = D f(x1,x2) = 2.x1 + 5x2 Dx1
Step 2: f1(x1,x2) = D f(x1,x2) = 0 => 2.x1 + 5x2 = 0 Dx1 => x1 = 5x2 2
Step 3: Critical points of function f(x1,x2) = > if x2 = 1 then x1 =5/2 and if x2 =
2 then x1 =5. So, we assume we have critical points (1,5/2) and (2, 5)......etc
Step 4: Then we calculate second order derivative of function f(x)  f11(x1) = D
f1(x1) = 2 + 5 x2 Dx1
Learn More About Rational Numbers Additive Inverse Worksheet
Step 5: At last we calculate value of f11(x1) for critical points  [f11(x)](1,5/2) =
2 + 5x2 = 2 + 5(5/2) = 2  25/2 = 21/2 (1,5/2) produces negative value, so
these critical points are called as a local maxima.
[f11(x)](1,5/2) = 2 + 5x2 = 2 + 5(5) = 2 + 25 = 27 (5,2) produces positive value
so, these critical points are called as a local minima.
ThankYou
TutorCircle.com
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