This is Advanced calculus of several variables. In detailed explanation please. Thank you.

I was able to get the x=y=z=10 and lambda=2/5.

Using H=2xy+2xz+2yz-(2/5)xyz+1000(2/5), I found the 2nd partial derivatives to make the matrix but all my 2 are positive.

1) How come the 2s in matrix are negative?

2) How do I get q(xyz)= - 2xy - 2xz - 2yz?

3) How do I find v1=(1, -1, 0) and v2=(1, 0, -1)?

I understood why v=sv1+tv2=(s+t, -s, -t), but I am not sure how to get q(v)=2s^2+2st+2t^2 by substituting this info to (10)

4) How do I get q(v)=2s^2+2st+2t^2?

Can someone actually show me this example step by step?

\f154 Il Multivariable Differential Calculus Vom (a) VG (a) The translate of 1 to a M Figure 2.42 We will obtain sufficient conditions for f to have a local extremum at a by considering the auxiliary function H : " - # for f at a defined by H(x) = f(x) - > 2, G.(x). (8) Notice that (7) simply asserts that V/(a) = 0, so a is an (ordinary) critical point for H. We are, in particular, interested in the quadratic form = = [ D, D, H(a)h,h; of H at a. Theorem 8.9 Let a be a critical point for fon M, and denote by q : "- the quadratic form at a of the auxiliary function H =f- ) , 1, G,, as above. If f and G are both of class 4 in a neighborhood of a, then f has (a) a local minimum on M at a if q is positive-definite on the tangent space T, to M at a, (b) a local maximum on A at a if q is negative-definite on T. , (c) neither if q is nondefinite on T. . The statement. that "q is positive-definite on T, ," naturally means that q(h) > 0 for all nonzero vectors he T.; similarly for the other two cases.Example 4 In Example 4 of Section 5 we wanted to find the box with volume 1000 having least total surface area, and this involved minimizing the function f(x, y. z) = 2xy + 2xz + 2yz on the 2-manifold M c ' defined by the constraint equation g(x, y, Z) = XYZ - 1000 = 0. We found the single critical point a = (10, 10, 10) for f on M, with 1 = . The auxiliary function is then defined by h(x, y, z) = f (x, y, z) - Ag(x, y, Z) = 2xy + 2xz + 2yz - 3xyz + 400.We find by routine computation that the matrix of second partial derivatives of h at a is ONN NO so the quadratic form of h at a is q(x, y, z) = -2xy - 2xz - 2yz. (10) It is clear that q is nondefinite on 3. However it is the behavior of q on the tangent plane T, that interests us. Since the gradient vector Vo(a) = (100, 100, 100) is orthogonal to M at a = (10, 10, 10), the tangent plane , is generated by the vectors V1 = (1, - 1, 0) and V2 = (1,0, - 1). Given ve T. , we may therefore write V = SV, + 1V2 = (s + 1, -s, -1). and then find by substitution into (10) that q(v) = 2s' + 2st + 212. Since the quadratic form s + st + 12 is positive-definite (by the ac - b test), it follows that q is positive-definite on T. . Therefore Theorem 8.9 assures us that f does indeed have a local minimum on M at the critical point a = (10, 10, 10)