MAX/MINS with multivariables

[Giuseppe]

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?

 

[Nate810]

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.

 

[quicknote]

Your approach seems to be correct. However, it is important to note that while finding critical points (where the gradient is equal to 0) is a necessary condition for max/min, it is not sufficient. In other words, just because the gradient is 0 at a certain point does not guarantee that it is a max/min.

In order to determine if the critical points you found are actually max/min, you need to use the second derivative test. This involves taking the second partial derivatives and plugging in the values of x and y at the critical points. If the second derivative test is inconclusive, then you may need to consider other methods such as using Lagrange multipliers.

Additionally, when parameterizing the equation, make sure to specify the range of t. In this case, t should range from 0 to 2π in order to cover the entire circular disk.

Overall, your approach is correct but you may need to do some additional calculations to determine if the critical points are actually max/min.