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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 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.


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Maxima and Minima of Functions of Several Variables