1: Let f(x,y) = (x^2 - 1)(y^2 - 1). Find and classify the...

1: Let f(x,y) = (x^2 - 1)(y^2 - 1). Find and classify the critical points of f. f(x,y)=x^2y^2-x^2-y^2+1 CP: (1,1)(1,-1)(-1,1)(-1,-1) f_x = 2xy^2 - 2x f_xx = 2y^2-2 D = f_xx ? f_yy - f_xy f_y = x^2 2y - 2y f_yy = 2x^2-2 D(1,1)=0-4=-4<0 saddle f_xy = 4xy D(1,-1)=(0-0)+4=4>0 max/min 2xy^2-2x=0 x^2 2y - 2y = 0 D(-1,1)=0+4=4>0 max/min 2xy^2=2x 2x^2y=2y D(-1,-1)=0-4=-4<0 saddle y^2=1 y=±1 x^2=1 x=±1 2: Use Lagrange multipliers to find (x,y) inputs that optimize f subject to the constraint x^2 + y^2 = 2. gx=2x gy=2y 2xy^2-2x = ?2x (?+1)+(?+1)=2 (2xy^2)/2x = (?2x+2x)/2x y^2 = ? + 1 2x^2y-2y = ?2y 2x^2y = ?2y+2y x^2 = ? + 1 3: What are the maximum and minimum values of f over the domain x^2 + y^2 ? 2?

Transcript

00:01 In this question, the function f of x, y that is equals to x square minus 1 into y square minus 1, it is given.

00:14 Now, to find critical points, to find critical point, quote, partial derivative of the function with respect to x equals to 0 and partial derivative of the function with respect to y equals to 0.

00:34 So, here partial derivative of the given function with respect to x that will be 2 x into y square minus 1 and this is equals to 0.

00:48 So, this implies 2 x y square minus 1 equals to 0.

00:53 So, from here x equals to 0 and y equals to plus minus 1.

01:01 Now, partial derivative of this function with respect to y, so that is equals to x square minus 1 into 2 y.

01:11 Now, this will be equals to 0.

01:14 So, x square minus 1 into 2 y equals to 0.

01:18 This implies y equals to 0 and x equals to plus minus 1.

01:25 Therefore, the critical points, therefore the critical points of, critical points of f of x comma y are 0 comma 0, 0 comma 1, 0 comma minus 1, 1 comma 0, 1 comma minus 1, 1 comma 1 and minus 1 comma 0, minus 1 comma 1.

02:02 And when x is negative, y is negative both.

02:06 Now, to classify, so here, here derivative of, partial derivative of this function with respect to, with respect to x again, so that will be equals to, so del f by del x square.

02:23 So, partial derivative of this function with respect to x again that will be twice of y square minus 1 and this is represented by f of x, x.

02:38 Then, partial derivative of the function with respect to y again, so that will be equals to with respect to y again.

02:49 So, here this is partial derivative with respect to y.

02:52 So, y again this will be twice of x square minus 1 which is represented by f of y, y and a del square f, del square f upon del x del y.

03:10 So, del x and del y, so that will be equals to partial derivative of this function with respect to y.

03:19 So, with respect to y, this will be with respect to y, so here f x, y.

03:27 So, that will be twice of x and this will be 2 y.

03:32 So, this is equals to 4 x, y.

03:35 So, from here f x, y, this is equals to f x, y and this is equals to 4 x, y.

03:43 Now, the critical points are to classify critical points, to classify critical, classify critical points.

03:57 So, here value of at 0, 0.

04:01 So, first critical point is at 0, 0...

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