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Problem 7.3. Let f (x, y) = x6 + 3xy + y2 + y4.
(a) Show that f remains unchanged if you replace x by ?x and y by?y. Hence,
if (x, y) is a critical point of f, so is (?x, ?y). Thus, criticalpoints other than
(0, 0) come in ± pairs.
140 7 Optimization in Several Variables
(b) Compute the partial derivatives of f . Show that solve applieddirectly to
the system fx = fy = 0 fails to locate any of the critical pointsexcept for (0, 0).
(c) Let’s compensate by eliminating one of the variables and thenusing solve
followed by double. First solve for y in terms of x in the equationfx = 0.
Substitute back into the formula for fy and then apply first solveand then
double. You should end up with three critical values of x, giving atotal of
three critical points. Find the numerical values of theircoordinates. (Be sure
you have set x and y to be real; otherwise you will also end upwith many
irrelevant complex critical points.)
(d) Confirm the calculation of the critical points by graphing theequations fx =
0 and fy = 0 on the same set of axes (using fimplicit and hold on).You
should see exactly one additional pair of critical points (in thesense of (a)).
(e) Classify the three critical points using the second derivativetest.
(f) Apply fminsearch to f with the starting values (1, 1) and (0,0). Show
that in the first case you go to a minimum and that in the secondcase you stay
near the saddle point.