Maths

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Maths

Devika
Let the domain of the function h be the set of all (x,y) satisfying both 1 <= x <= 2 and 0 <= y <= x -1.
And let h(x,y)= x^2y(x-y-1). Find the global extreme points.
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Re: Maths

Sinistral
Is it h(x,y) = (x^2)*y(x-y-1) ? or h(x,y)=(x^2y)*(x-y-1)?
I am trying to solve it assuming former to be right.
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 "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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Re: Maths

Devika
its x^2 *y (x-y-1)
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Re: Maths

Sinistral
In reply to this post by Devika
we need to find extreme points of h(x,y) in the given domain.

so ∂h/∂x = h_x = xy(3x-2y-2)
similarly ∂h/∂y= h_y= x^2(x-2y-1)

equating the above 2 to zero to get the critical points.
(x_c,y_c)≡(.5,-0.25)  which is outside the given domain.

so we will try to check the maximum/minimum values of  h(x,y) at its boundary.
the 3 boundaries are x axis (from x=1 to 2)
                             x=2 (from y=0 to 1)
                           y=x-1 (from x=1 to 2)
on x axis (from x=1 to 2) a general point is (k,0). 1<=k<=2. it makes h(x,y) =0
on y=x-1 (from x=1 to 2) a general point is (t,t-1). 1<=t<=2. it makes h(x,y) =0
on  x=2 (from y=0 to 1) a general point is (2,k) st 0<= k <=1. so we get h (k)= -4k^2 + 4k
h(k) is maximum at k=1/2. so h_max=1
point of maxima ≡ (2,0.5)

now it is obvious that h(x,y) >=0. so h_min =0
point of minima lies on x axis (from x=1 to 2) & y=x-1 (from x=1 to 2).

These days I am making lot of mistakes. I hope this time I didn't make one .
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 "You don't have to believe in God, but you should believe in The Book." -Paul Erdős