Calculus distance maximizing 
Point $(x, y)$ satisfies the inequality $x^4 + y^2 \leqslant 1$. Determine the largest possible distance from the origin for $(x, y)$.

So the largest distance will result when $x^4+x^2=1$.
The distance from the origin for point $(x,y)$ is $d=\sqrt{x^2+y^2}$, but from the given inequality we see that $y^2=1-x^4$ therefore we get $d(x)=\sqrt{x^2+1-x^4}$. I know that in order to maximize distance we should find the derivative of $d(x)$, but it seems rather peculiar to differentiate that expression. Is there another way I could go with this?
 A: Let $d(x,y)=x^2+y^2$. You're after the maximim of $d$ in the region $\{(x,y)\in\Bbb R^2\mid f(x,y)\leqslant1\}$, where $f(x,y)=x^4+y^2$. That maximum cannot be attained in the region $\{(x,y)\in\Bbb R^2\mid f(x,y)<1\}$ because that's an open set and therefore if the maximum was attained there, it be attained at a critical pont of $d$. But the only critical point of $d$ is $(0,0)$ and $d$ has actually a minimum there.
So, you can use the method of Lagrange multipliers to find the maximum in the region  $\{(x,y)\in\Bbb R^2\mid f(x,y)=1\}$. To do so, you solve the system$$\left\{\begin{array}{l}\frac{\partial d}{\partial x}(x,y)=\lambda\frac{\partial f}{\partial x}(x,y)\\\frac{\partial d}{\partial y}(x,y)=\lambda\frac{\partial f}{\partial y}(x,y)\\f(x,y)=1\end{array}\right.\iff\left\{\begin{array}{l}2x=4\lambda x^3\\2y=2\lambda y\\x^4+y^2=1.\end{array}\right.$$Its solutions are $(\pm1,0)$, $(0,\pm1)$, $\pm\left(\frac{\sqrt2}2,\frac{\sqrt3}2\right)$ and $\pm\left(\frac{\sqrt2}2,-\frac{\sqrt3}2\right)$. The value that $d$ takes at the first four of these points is $1$, in the remainig ones it is $\frac54$.
Therefore, the minimal distance is $\frac{\sqrt5}2$.
A: Differentiating the square of $d$, we get $2x-4x^3=0\implies x=0\lor \pm\sqrt2/2$.
Those are the critical points, and we get $1$ and $\sqrt5/2$ when we evaluate.
So the maximum distance is $\sqrt5/2$.
A: Attempt:
Constraint: $x^4+y^2 \le 1$;  $y^2\le 1-x^4;$
Maximize : $x^2+y^2\le x^2+1-x^4=$
$-(x^4-x^2)+1=$
$-[(x^2-1/2)^2-1/4]+1=-(x^2-1/2)^2+5/4\le 5/4$.
Maximum attained at $x^2=1/2$; 
$y^2=5/4 -1/2=3/4.$
A: Let $P(a,b)$ be the point on $x^4+y^2=1$. At the maximal distance from the origin $O$, $OP$ is normal to the curve, i.e. 
$$y’= -\frac{2a^3}{b}=-\frac ab  \implies a^2=\frac12,\>\>\>b^2=\frac34$$
Thus, the maximal distance is 
$$\sqrt{a^2+b^2}=\frac{\sqrt5}2$$
