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Find all of the critical points. Question 1b in the above pic. So fx(x,y) = 6xy-

ID: 2834779 • Letter: F

Question

Find all of the critical points. Question 1b in the above pic.                 


                    So fx(x,y) = 6xy-6x                 

                    and                 

                    fy(x,y) = 3y^2+3y^2-6y                 


                    Please show work in setting these equal to zero and finding all of the critical points.

Find all of the critical points of given function, There use the Second Derivative Test. Let f(x, y, z) - 2ex 2 - v2 Compute of f Find the directional derivative of f at (1,-1.2)in the direction (2,2,1)

Explanation / Answer

fx(x,y) = 6xy-6x=6x(y-1)

There is a typo on fy(x,y)

You mean 3x^2 + 3y^2 - 6y

From the first expression, we get that either x = 0 or y = 1

You plug each of these into the second expression.

If x = 0, we have 3y^2 - 6y = 0, or 3y(y-2) = 0

This has solutions y = 0 and y = 2

Thus, we have critical points (0, 0) and (0, 2)

From y = 1, we have 3x^2 +3*1 - 6*1 = 0

3x^2 = 3

x = plus or minus 1

Thus, we have critical points (1, 1) and (-1, 1)

Thus, we have four critical points, (-1, 1), (0, 0), (0, 2), (1, 1)

Next, we look at the second derivtive

We have fxx = 6y - 6; fxy = fyx = 6x; fyy = 6y - 6

Thus, at (-1, 1), we have

0 -6

-6  0

Think of the function -6xy at (0, 0). This has the same Hessian. Thus, we see that this is a saddle point.

At (0, 0), we have

-6 0

0   -6

This is negative definite, (2 eigenvalues of -6), so we hae a maximum.

At (0, 2), we have

6   0

0   6

This is positive definite (2 eigenvalues of 6), so we have a minimum

Finally, (1, 1)

We have

0 6

6 0

This is again a saddle point.

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