Find the absolute maximum and minimum values offon the setD.

f(x, y) =xy^2+ 9D= {(x, y) |x≥ 0,y≥ 0,x^2+y^2≤ 3}

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- Mar 17th 2009, 07:12 PMacg716Maximum and Minimum of multivariable calc
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} - Mar 17th 2009, 08:22 PMmatheagle
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. - Mar 17th 2009, 08:34 PMacg716
So the minimum is when x and y are both -sqrt(3)?

I'm a bit confused. - Mar 17th 2009, 08:42 PMmatheagle
- Mar 17th 2009, 09:15 PMacg716
sorry i meant maximum!