Thread: Absolute Max and Min

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 $\displaystyle y^2 = 3-x^2$ and plugging that in to the original to get $\displaystyle g(x) = f(x,y) = -x^3+3x+7$. 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.

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???

I'm guessing that he/she means the partial derivatives by fx and fy.
BUT they are missing powers here.
I'm guessing f= xy^2 and the constraint is the interior of the circle x^2+y^2 =3.
Otherwise I have no idea what x2 and y2 are.

If the function under study is...

$\displaystyle f(x,y)= x \cdot y^{2}$ (1)

... [we are omitting the constant term + 7...] the partial derivatives are effectively...

$\displaystyle f_{x}(*,*)= y^{2}$

$\displaystyle f_{y}(*,*)= 2\cdot x\cdot y$ (2)

Both the f(*,*) and the derivatives are non negative in the quarter of circle $\displaystyle [x\ge 0, y\ge 0, x^{2} + y^{2} \le 3]$, so that the maximum is on the circular contour, i.e. a point in which is...

$\displaystyle y^{2}= 3 - x^{2}$ (3)

Substituting (3) in (1) we obtain the function of the only x...

$\displaystyle f(x,3-x^{2}) = 3\cdot x - x^{3}$ (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...

$\displaystyle [x,y]= [1,\sqrt{2}]$

... and its value is $\displaystyle f(1,\sqrt{2})= 2 [+ 7]$...

Since on the x and y axes is f(*,*)=0, the (1) has no minimum points...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$