I assume you mean F(x,y) = x^3 - y^2 - 3x + 6y.
dF/dx = 3x^2 - 3, so dF/dx = 0 when x=1,-1
dF/dy = -2y + 6, so dF/dy = 0 when y=3.
Therefore, the critical points are (1,3) and (-1,3).
Fxx = d^2F/dx^2 = 6x
Fyy = d^2F/dy^2 = -2
Fxy = Fyx = d^2F/dxdy = 0
Therefore, the Hessian is the matrix
[ 6x 0
0 -2 ]
which is negative definite at (-1,3), so that point is a relative maximum. At (1,3), the matrix is indefinite, so it's a saddle point.
Note: the Hessian is
a) positive definite (minimum) if Fxx*Fyy - Fxy^2 > 0 and Fxx, Fyy > 0,
b) negative definite (maximum) if Fxx*Fyy-Fxy^2 > 0 with Fxx, Fyy < 0,
c) indefinite if Fxx*Fyy - Fxy^2 < 0
This is probably the test you're more familiar with, but is equivalent to what I described. When Fxy = 0, then it's enough to check whether the diagonal elements are both positive, both negative, or of different signs to classify a critical point.
2006-10-24 10:47:12
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answer #1
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answered by James L 5
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Couple of diverse strategies, yet i'm assuming those equations are correct, so which you are able to use a gadget to sparkling up them. Set them up in a gadget. 3x + 6y = 30 x + 6y = 20 -or- [ 3 6 | 30] [ a million 6 | 20 | 30] [ a million 3 6 | 30] [ a million 6 | 20 20] <---- matrix notation Subtracting row 2 from row a million provides here: 3x + 6y = 30 -x - 6y = -20 ------------------- 2x + 0y = 10 Divide 10 by way of 2 to get x = 5. Plug that x fee lower back into the two of the unique equations to get y = 15/6
2016-11-25 02:36:44
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answer #2
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answered by baksi 3
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