Question 1: Calculate the optimal points of the following...

Question 1: Calculate the optimal points of the following multivariate functions. a) $y = f(x_1, x_2) = x_1 x_2^2 + x_1^3 x_2 - x_1 x_2$ b) $y = f(x_1, x_2) = x_1^4 + x_1^2 - 6x_1 x_2 + 3x_2^2$ c) $y = f(x_1, x_2) = x_1^2 + x_2^2 + 3x_1 x_2$ d) $y = f(x_1, x_2) = -x_1^2 + 6x_1 - 4x_2^2 + 16x_2$ e) $y = f(x_1, x_2) = 3x_1^4 + 3x_1^2 x_2 - x_2^3$

Transcript

00:01 In this question, the function x1 x2 is defined by x1 x2 whole square plus x1 cube x2 minus x1 x2.

00:16 Now its partial derivative with respect to x1 will be equals to x2 square and here it will be thrice of x1 square x2 and minus x2 and from here taking x2 common so del f upon del x1 will be equals to taking x2 common here it will be x2 plus 3 times of x1 square and here it will be minus 1.

00:53 Now this is equation number 1.

00:56 Now the partial derivative of the function with respect to x2 will be equals to that will be equals to x1 thrice of x2 plus x1 cube and here it will be 1 and minus x1.

01:17 So in this case taking x1 common so del f upon del x2 equals to x1 thrice of x1 plus x1 square minus 1.

01:33 Now this is equation number 2.

01:35 So for optimal point so for optimal point so partial derivative with respect to x1 and y1 will be 0.

01:49 So del f upon del x1 del x1 equals to 0 and del f upon del x2 equals to 0.

02:01 So here del f upon del x2 that will be first of all here it will be x2 then x2 plus 3 x1 square minus 1 equals to 0 and here it will be x1 x1 twice of x2 plus x1 square minus 1 equals to 0.

02:27 So this implies x2 equals to 0.

02:31 So from here x2 equals to 0 or x2 plus thrice of x1 square minus 1 equals to 0.

02:41 So this implies x2 equals to 0 or thrice of x1 square minus 1 thrice of x1 square minus 1 equals to 0 because for x2 equals to 0 here it will be 0.

03:00 So this implies x1 equals to plus minus 1 upon root 3.

03:08 So from here we get x2 equals to 0 and x1 equals to plus minus 1 by root 3.

03:19 Then after this if x2 equals to 0 if x2 equals to 0 then here del f upon del x2.

03:33 So del f upon del x2 is equals to x1 2 x2 plus x1 square minus 1.

03:42 So here if x2 equals to 0 x2 equals to 0.

03:48 So here x1 twice of x2 plus x1 square minus 1 this is equals to 0...

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