Maximum and minimum values of $x + y$ given that $~~x^2+4y^2 \leq 1$.

Question

Find the maximum and minimum values of $~x + y~$ with respect to the constraints $~~x^2+4y^2\leq 1.$

My Attempt:
Let us assume $~~f(x,y)=x+y~~$ and $~~g(x,y)=x^2+4y^2-1.$ Now Consider a function $~~L(x,y,k)=f+kg.$
Now notice that \begin{align*} \frac{\partial L}{\partial x}=0 \implies & 1+2xk=0 \\ \frac{\partial L}{\partial y}=0 \implies & 1+8yk=0. \end{align*} Now from here I can conclude $~~x,y,k \neq 0.$ But I am not able to find the maximum and minimum values of $~~f$.

Answer 1

Your approach is valid on the boundary. But you're right, there are no critical points in the interior.

Now you know $2xk=-1=8yk$. Since $k\neq 0$, we get $x=4y$. With $x^2+4y^2=1$ you get $(4y)^2+4y^2=1$.

Answer 2

Substitute $x=a\cos\phi$ and $2y=a\sin\phi$ for some $|a|\le1$. We just need to find maximum and minimum values of $a\cos\phi+\frac a2\sin\phi$.

Maximizing with respect to $\phi$ gives $\pm\sqrt{a^2+\frac{a^2}{4}}=\pm a\sqrt{\frac54}$. Taking $a=\pm1$, we get maximum and minimum to be $\pm\frac{\sqrt5}2$.