Find the absolute maximum and minimum values of f on the set D.
f(x, y) = xy^2 + 9 D = {(x, y) | x ≥ 0, y ≥ 0, x^2 + y^2 ≤ 3}
Find the absolute maximum and minimum values of f on the set D.
f(x, y) = xy^2 + 9 D = {(x, y) | x ≥ 0, y ≥ 0, x^2 + y^2 ≤ 3}
I think you can just use common sense here.
The 9 does NOT affect the min and the max.
I now see that $\displaystyle x\ge 0$ and $\displaystyle y\ge 0$,
so the min is 9 which occurs along the x axis and the y axis.
Next insert the constraint in the function to get the max along the circle in the first quadrant.
$\displaystyle f(x)=xy^2+9=x(3-x^2)+9=3x-x^3+9$.
Now maximize this wrt x and you get x=1, since we're only in the first quadrant.
That gives $\displaystyle y^2=2$ and a max of 11.
I also got this via LaGrange, to check my work.