Evaluating limit for Penalty method: (x-1)²[y²-(x-1)³] = x /3K. y[y²-(x-1)³] = -y/2K

[keerthanbantwal]

Hi, I came across the following limit evaluation in one of the optimization textbooks:

Taking limit as K tends to ∞ , we get x = 1, y = 0
(x-1)²[y²-(x-1)³] = x /3K
y[y²-(x-1)³] = -y/2K

Can anyone please help me figure out how the limit was evaluated?

An additional probably useful information would be:
y²-(x-1)³ = 0

Thanks in advance.

 

[topsquark]

I think you might be making this too hard on yourself. As K tends toward infinity the RHS of each expression goes to 0. Then we have
[math](x - 1)^2(y^2 - (x - 1)^3) = 0[/math]and
[math]y(y^2 - (x - 1)^3) = 0[/math]
The first equation then becomes [math](x - 1)^2 = 0 \text{ or } y^2 - (x - 1)^3 = 0[/math].

The second equation then becomes [math]y = 0 \text{ or } y^2 - (x - 1)^3 = 0[/math].

I leave the rest to you.

-Dan

 

[Steven G]

Hmm, if y²-(x-1)³ = 0, then why is x = 1, y = 0 the only solution????

 

[keerthanbantwal]

I think you might be making this too hard on yourself. As K tends toward infinity the RHS of each expression goes to 0. Then we have
[math](x - 1)^2(y^2 - (x - 1)^3) = 0[/math]and
[math]y(y^2 - (x - 1)^3) = 0[/math]
The first equation then becomes [math](x - 1)^2 = 0 \text{ or } y^2 - (x - 1)^3 = 0[/math].

The second equation then becomes [math]y = 0 \text{ or } y^2 - (x - 1)^3 = 0[/math].

I leave the rest to you.

-Dan

Thank you for the reply.

Yes, that was my first impression. But y²-(x-1)³ = 0 is a constraint and can never be false. So, I assumed the there has to be another way to evaluate such limits. But I believe that has to be to the logic.

 

[keerthanbantwal]

Hmm, if y²-(x-1)³ = 0, then why is x = 1, y = 0 the only solution????

It was an optimization problem minimizing the function x² + y² . Subjected to the constraint. The Lagrange method fails in this case. Hence penalty method was demonstrated.

 

[Steven G]

Why not just give us the entire problem so we know what we are working with??