Lagrange Multiplier Optimisation: max/min dist of g(x,y) = (x^6)+(y^6)=1 to origin

[Ovosch]

Hello. My question is:
How to find the max/min distance from the curve g(x,y) = (x^6)+(y^6)=1 to the origin using Lagrange Multipliers.

I get stuck because I get lambda = 0

here is my solution:

distance from the origin √f(x,y) = √[(x-0)^2 + (y-0)^2] f(x,y) = x^2 + y^2

∂f/∂x = 2x ∂f/∂y = 2y ∂g/∂x = 6x^5 ∂g/∂y = 6y^5 ( µ = lambda )

2x = µ*6x^5 2y = µ*6y^5 simplify and plug into g(x,y) => (µ*3x^5)^6 + (µ*3y^5)^6 = 1 ;
take the factors out => [(3µ)^6] * [(x^30)+(y^30)] = 1

if i plug in the values of the origin (0,0) then I get (3µ)^6 * 0 = 1 so my µ (or lambda) is 0

Please help me out.
Thank you in advance.

 

[ksdhart]

I follow you up until the step where you say "simplify and plug into g(x,y)." The following is what I believe your steps were:

\(\displaystyle 2x=\lambda\cdot6x^5\)

\(\displaystyle x=\lambda\cdot3x^5\)

And similar for y, then you plugged these values of x and y into your g(x,y). But, if that's the case, your process isn't correct. If you're solving for x, you can't end up with x terms on both sides of the equation. You need to divide through by x^5 to have x terms only on one side:

\(\displaystyle \frac{x}{x^5}=\lambda \cdot 3\)

\(\displaystyle x^{-4}=\lambda \cdot 3\)

As you can see, this method can be done, but it will get ugly. Instead, I would solve for lambda. Once you get an expression for lambda in terms of x and another expression in terms of y, set them equal. What do you notice? What relationship between x and y can you extrapolate? Then you have an expression for x in terms of y (or vice versa), so plug it into g(x,y) and evaluate.

 

[Ovosch]

Thank you for your response.

If I do the maths correctly, I suppose i should get

λ = 1/(3x⁴) = 1/(3y⁴​) therefore x = y

so if I plug this into the equation of the curve i get x⁶ + x⁶ = 1 => x⁶ = 1/2 x = +1/(2)^1/6 V -1/(2)^1/6
and same for y. If so then all the values are the same, whereas i should get the max and the min values for the distances from the origin to the curve, therefore this answer is wrong.

P.S. The graph basically looks like a square with rounded corners. So i should probably get ~1 for my minimum value and ~√2 for my maximum value.

 

[ksdhart]

Hmm.. that is interesting. I can't find anything wrong with the process, yet when I put aside the Lagrange Multipliers and just think about it logically, I do agree with your analysis. For the general curve x^n+y^n where n is an even integer, the points farthest from the origin will lie on the diagonals, and their distance is the nth root of 2 [or 2^(1/n)]. Similarly, the closest points to the origin will lie on the coordinate axes [(0,1), (1,0), (-1,0), and (0,-1)]. It seems I'm a bit out of my league here, and sadly I'm without an answer as to why the Lagrange Multipliers aren't bearing out the correct answers. Sorry.