course Mth 275
Section 12.7This section is mostly very clear. Some of it (especially the modeling problems) seem to be an extension of linear programming to R3 and non-linear functions?articularly the identification of endpoints as possible extrema. I had no difficulty with most of the problems. In number 21 [finding and classifying the critical points of f(x,y) = x^2+y^3+768/(x+y)], I was able to identify and classify (6,2), but I could not come with the second point the book gives: -8.99,-2.45), even working backward knowing the point.
f_x = 2 x - 768 / (x + y)^2 and f_y = 3 y^2 - 768/(x+y)^2.
These are zero when 2 x ( x + y)^2 = 3 y ^2 ( x + y)^2 = 768.
The only possible solutions are when 2 x = 3 y^2 (i.e., since 2 x ( x + y)^2 = 3 y ^2 ( x + y)^2 and (x + y)^2 is zero only when x = y, we can divide through by (x + y)^2).
Substituting x = 3/2 y^2 we thus obtain
2 ( 3/2 y^2) ( 3/2 y^2 + y)^2 - 768 = 0, or expanded
27·y^6/4 + 9·y^5 + 3·y^4 - 768 = 0.
Expanding we get a sixth-degree polynomial in y which is equal to 0. We wouldn't expect to get an exact solution, but it turns out that against expectations y = 2 is a solution so x = 3/2 (2^2) = 6 gives us the point (2, 6).
Dividing 27·y^6/4 + 9·y^5 + 3·y^4 - 768 by (y - 2) then gives us a fifth-degree polynomial and tells us that the other solution is the solution to
9·y^5 + 30·y^4 + 64·y^3 + 128·y^2 + 256·y + 512 = 0.
We could attempt to factor this using the rational-roots theorem, etc., but at this point it's OK to just graph the thing and approximate the zero, which is at y = -2.45.
DERIVE gives the exact solution as y = (3896/729 - 32·‹183/81)^(1/3) - (32·‹183/81 + 3896/729)^(1/3) - 2/9, but we just don't want to get that involved here.