optimize $x^2+y$ on the constraint that $x^2-y^3=0$

Question

Problem: Find maximum and minimum value for the function $f(x,y)=x^2+y$ on the constraint $x^2-y^3=0$.

My Solution:

I have started solving this by using the Lagrange method to find points where the gradients are perpendicular and where the points are also satisfied by the constraint.

I get (by solving for when the determinant that holds the two gradients is zero) the equation $-2x(3y^2+1)=0$.

So this gives me first with $-2x=0$ and the constraint $x^2-y^3=0$ the system of equations: $$x=0$$ $$x^2-y^3=0$$ Which gives me the point $(0,0)$ with value $f(0,0)=0$

With $3y^2+1=0$ and the constraint I get: $$y^2=-1/3$$ $$x^2-y^3=0$$

Questions:

• My question for the second system of equation is: with $y^2=-1/3$ does this mean that there is no solution? Or did I do something wrong?

• The constraint is not a compact set of value, it is not bounded? My textbook gives the answer to this problem as minimum value is 0 and maximum does not exist. How can I prove this and how do I know that 0 is the minimum and not the maximum value?

Answer 1

WLOG $x=t^3,y=t^2$

$x^2+y=t^6+t^2=t^2(t^4+1)\ge0$