Find Absolute Max/Min of f(x,y)=xy^2 w/ Domain x^2+y^2≤4

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The discussion focuses on finding the absolute maximum and minimum of the function f(x,y) = xy² within the domain defined by x² + y² ≤ 4. Participants emphasize the importance of calculating partial derivatives, setting them to zero to find critical points, and evaluating the Hessian matrix for classification. Additionally, they highlight the necessity of examining boundary points, specifically those on the circle of radius 2, to determine extrema. The consensus is that extrema can occur at both critical points and boundary points, necessitating a thorough evaluation of both.

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der.physika
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Can anyone tell me the general procedure in doing the following procedure?

f(x,y)=xy^2 with domain x^2+y^2\leq4

Find it's absolute max & absolute min.

Okay, here is my thought procedure, tell me what I can fix.

So I would basically say, find the partial derivatives with respect to x and y and set them equal to zero.

f_x=y^2=0 f_y=2yx=0

so what's up? I plug that into the original equation? and then do the whole matrix thing to find if it's an absolute max or min? so point (x,y)=(0,0)

Plug into the matrix \left(\begin{array}{cc}f_x_x&f_x_y\\f_x_y&f_y_y\end{array}\right)

But I don't know how I would go about considering the x^2+y^2\leq4, do I find the boundary point? What are those? (x,y)=(2,0)=(0,2)=(-2,0)=(0,-2) and then plug it into the original equation and then use

Plug into the matrix \left(\begin{array}{cc}f_x_x&f_x_y\\f_x_y&f_y_y\end{array}\right)

Am I on the right track? Can someone show me some guidance?
 
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You use the Hessian matrix (what you did above) to classify critical points. But the extrema do not have to occur at critical points--they can also occur at the boundary. In this case, the boundary consists of a circle of radius 2. Think about how the function behaves on this circle...maybe rewrite in terms of angle and see what you find.
 
In addition to what TinyBoss told you, you don't really need the Hessian in this problem. You first locate the critical points in the interior (there are lots of them,no?). Once you figure out the extremes on the boundary, which don't necessarily have to be at the axis intercepts just because you like them, you just list the points in the original function xy2. Then you can eyeball them to see the absolute max and min.
 

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