Is My Approach to Finding Max/Min on a Circular Disk Correct?

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The discussion centers on finding the maximum and minimum values of the function f(x,y) = xy^2 over the circular disk defined by x^2 + y^2 ≤ 3. The user correctly identifies the need to compute the gradient and evaluate critical points, starting with the gradient calculation. However, the user makes an error in the gradient derivative, which should be <y^2, 2xy>, leading to incorrect critical points. Correcting the derivative yields four critical points, with two representing maxima and two representing minima. The user seeks clarification on their approach and is guided to rectify their derivative calculation for accurate results.
Giuseppe
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Hey I was wondering if anyone can tell me if i am doing this right.

f(x,y) = xy^2 ; R is the circular disk x^2+y^2<=3

So first i took the gradient, since I know a mix/min can exist if the gradient is equal to 0.

Gradient of X= y^s
Gradient of Y= 2xy

So the point (0,0) can be considered right?

Anyway, I know I have to test region edges, so I parameterized the equation.

r(t) = <radical 3 cos(t),radical 3 sin(t)>

i plugged those values of x and y into my equation , and then took the derivative.

After some simplification, I came up with

sin(t)(3*radical3*cos(t)^2-3*radical3)

t= 0, pi, pi/2, 3pi/2 (right?)

so i found the x and y value when t is equal to those values

in conclusion, i have these points.

(0,0)
(0,radical3)
(0,-radical3)
(radical3,0)
(-radical3,0)

I tested these values in the equation xy^2 = f(x,y)

and found that there is no max or min...i don't think this is right.

Can someone help me find my mistake?
 
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Your mistake is in the derivative you took. The gradient of f(x,y) should be <y^2,2xy>. You should have gotten sin(t)(3*radical3*cos(t)^2+3*radical3)when you took the derivative. Then, when you plug in 0, pi, pi/2, 3pi/2 for t, you should get four critical points, two of which are maxima and two of which are minima.
 
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