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KnightsPG
04-15-2002, 10:14 AM
This question comes from the Averbach/Mehta manual, practice test 1, #32.

We are given f(x) = x^3 + 6xy + y^3 + 3

What are all points at which f has a relative maximum?

I take the derivative simultaneously and set equal to 0 to find the critical points. I can get to the point where the critical points are (0,0) and (-2, -2), but I can't determine how to answer between...

a) (0,0)
b) (-2,-2)
c) (0,0) and (-2,-2)

Similarly, Practice Test #2, #37

Let f(x,y) = 2(y^2) - 15y + (x^2)y - 2xy

At which of the following critical points does f have a relative maximum?

I. (-3,0)
II. (1,4)
III. (5,0)

I see how you take the derivative to get the critical points and I see how you take the second derivative to see if by inputing the points you can determine whether the value is less than 0. However, in this problem, we have two values that are less than 0, I and III. The back of the book gives the blunt answer f(-3,-0) is not an extremum of f. I don't understand what that means.

Can anyone help me out with these problems? Thanks!

Packet_Storm
04-15-2002, 11:45 AM
f(x) = x^3 + 6xy + y^3 + 3

Solve these for zero:
Fx = 3x^2 + 6y
Fy = 6x + 3y^2
thus giving you (a,b)

By now you should see that both x and y will be the same values. ie a = b.

Second Partials Test:
Fxx = 6x + 6
Fyy = 6y + 6
Fxy = 6

d = fxx(a,b)fyy(a,b) - [fxy(a,b)]^2

i) d &gt; 0 &amp; Fxx(a,b) &gt; 0, then rel. min
ii) d &gt; 0 &amp; Fxx(a,b) &lt; 0, then rel. max
iii) d &lt; 0, then (a, b, f(a,b)) is saddle pt.
iiii) d = 0, then inconclusive.

thanks goodness I just happen to have my calculus book handy.