Critical points of constrained optimization problem

Question

$f(x,y) = x^2+2y^2$ . Solve for critical points of f along the curve $h(x,y) = x^4+y^4-1=0$, where $x,y \in [0,1]$.

We set:
$\nabla f(x) = \lambda\nabla h(x)$
$h(x) = 0$

Above gives the equations: (1) $2x = 4\lambda x^3$, (2) $4y=4\lambda y^3$, (3) $x^4+y^4-1=0$.

***I solved the above equations using simultaneous equations, but it only gave one critical point for $x,y \in [0,1]$. There are supposed to be 3 critical points. Is there a simpler/better method to get critical points? (working is as below)
***Also, is there a systemic method to check what type of extrema this critical point is?

Simultaneous equations working:
Solving for x and y in terms of $\lambda$ and substitute it into the 3rd equation:
$x^4+y^4-1=0 \implies (\pm\frac{1}{\sqrt{2\lambda}})^4 + (\pm\frac{1}{\sqrt{\lambda}})^4-1 = 0 \implies \frac{1}{4\lambda^2}+\frac{1}{\lambda^2} = 1 \implies \lambda=\pm\frac{\sqrt5}{2}$
Hence, $x=\pm\sqrt\frac{1}{\sqrt5}$ and $\pm\sqrt-\frac{1}{\sqrt5}$, $y=\pm\sqrt\frac{2}{\sqrt5}$ and $\pm\sqrt-\frac{2}{\sqrt5}$.

Keeping in mind that $x,y \in [0,1]$, we ultimately have only one critical point: $x = \sqrt\frac{1}{\sqrt5}$ and $y=\sqrt\frac{2}{\sqrt5}$

Answer 1

Hints

• From your equation $(1)$, you only get $\ x^2=\frac{1}{2\lambda}\ $ if $\ x\ne0\ $. But $\ x=0\ $ is another solution of $(1)$, and this gives you another critical point.
• Likewise, $\ y=0\ $ is a solution of your equation $(2)$, and this gives you a third critical point.