you need both partials to be zero for relative extrema
Fx = y^2
Fy = 2xy
so clearly at (0,0) there is a min/max or saddle point
Also clearly, that will not be the absolute min/max, which will occur on the given circle.
y^2 <= 3-x^2
xy^2 <= x(3-x^2)
so, what is the absolute min of 3x-x^3 ?
that is when x=1, y=√2
Find the absolute maximum and minimum values of f on the set D.
f(x, y) = xy^2 + 5, D = {(x, y) | x ≥ 0, y ≥ 0, x^2 +y^2 ≤3}
absolute maximum value =
1 answer
1 answer