I tried doing this:
fx = y^2
fy = 2xy
critical point: (0,0).....even i don't understand what u have written.
what u have meant by fx,fy???
Hey i'm having difficulty solving this problem:
Find the absolute maximum and minimum values of f on the set D.
f(x, y) = xy2 + 7 D = {(x, y) | x ≥ 0, y ≥ 0, x2 + y2 ≤ 3}
I tried doing this:
fx = y^2
fy = 2xy
critical point: (0,0)
then i got and plugging that in to the original to get . setting that equal to zero i get x=1,-1. but i'm totally lost on what to do from here (or even if i've started correctly). Any and all help is greatly appreciated.
Thanks.
If the function under study is...
(1)
... [we are omitting the constant term + 7...] the partial derivatives are effectively...
(2)
Both the f(*,*) and the derivatives are non negative in the quarter of circle , so that the maximum is on the circular contour, i.e. a point in which is...
(3)
Substituting (3) in (1) we obtain the function of the only x...
(4)
... the derivative of which vanishes in x=0, x=-1, and x=1. But in x=0 is f(*,*)=0 and x=-1 is outside the quarter of circle, so that x=1 is the only solution. The maximum of f(*,*) in D is then in the point...
... and its value is ...
Since on the x and y axes is f(*,*)=0, the (1) has no minimum points...
Kind regards