This question Maximum with Lagrange multiplier Has no answer and I would be interested in knowing it.
The problem is the following :
You have to find the extremals of the function $$(x+y)^4+y^4 \mbox{ such that } x^4+y^4=1$$ One way could be to use the Lagrangian multiplier. But i can't solve the system. Any idea?
For now I've found : $$ 4(x+y)^3+4 a x^3 =0 $$ $$ 4(x+y)^3+4y^3+4 a y^3 =0 $$
By subtracting the two equations you got you end up with: $$4ax^{3}=4y^{3}+4ay^{3}$$ Dividing both sides by $4$, grouping terms and taking the cubic root you are left with: $$y=\sqrt[3]{\frac{a}{1+a}}x$$ Let's now call $k=\sqrt[3]{\frac{a}{1+a}}$ By substituting once again in the constraint you get: $$x^{4}=\frac{1}{1+k^{4}}$$ $$y^{4}=\frac{k^{4}}{1+k^{4}}$$ The initial function becomes: $$\frac{(k+1)^{4}}{1+k^{4}}+\frac{k^{4}}{1+k^{4}}$$ You can now minimize over $k$ and then compute $a$ from the definition