Optimizing with inequalities constraints

[Vuin]

Hi, i'm optimizing functions and have found no problems with equality constraints, but now i have to do them with inequalities constraint, what difference should i consider solving them? I'll paste what i did
Problem:
max 2y²-x
sub x²+y²≤1

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I started with writing the constraint like this
y²≤1-x²

i substituted it from the first function
2(1-x²)-x

I found the derivative and put it = 0 to find x
-4x-1
x=-1/4

having x i put that value in the constraint function to find y
y²≤1-x²
y²≤1-(-1/4)²
y≤±√15/4

I put the value of x and y in the first function to find the value of the max
x=-1/4 y=√15/4 z=17/8

Both x and y value are ±, so i did the same with x=1/4, since being y squared its value wouldn't change, and i get this
x=1/4 y=√15/4 z=13/8

To check if it's a max i found the determinant of the hessian matrix

0	-1/4	(151/2 )/4
-1/4	0	0
(151/2 )/4	0	4

Which is -1 <0, so a max

Is 17/8 the Max and 13/8 the Min? I feel like i'm missing something

 

[HallsofIvy]

You have the right answer but for the wrong reasons! Saying that \(\displaystyle x^2+ y^2 \le 1\) does NOT give \(\displaystyle y^2= 1- x^2\). You have gone from the inequality to the equation without reason.

The set of (x, y) such that \(\displaystyle x^2+ y^2\le 1\) is the disk of radius 1 centered at (0, 0). All that constraint means is that we have make sure any max or min is in that disk. The function to be minimized is \(\displaystyle 2y^2- x\). Its gradient is \(\displaystyle <4y, -1>\). That is NEVER equal to 0 so there is no critical point and, in particular, no max or min in the interior of the set. NOW you can say that since there is no max or min in the interior of the disk, they must be on the boundary and that is the circle \(\displaystyle x^2+ y^2= 1\).