Find the minimum value of $f(x,y) = x + y^2$ given the constraint $2x^2 + y^2 = 1$

Question

Find the minimum value of $x + y^2$ subject to the condition $2x^2 + y^2 = 1$.

1) I find $\nabla f$ and $\nabla g$ to get $$\nabla f(x,y) = (1, 2y) \\ \nabla g(x,y) = (4x, 2y)$$

Then I set up the system of equations

\begin{align} \nabla f(x,y) &= \lambda g(x,y) \\ 1 &= \lambda 4x \\ 2y &= \lambda 2y \\ 2x^2 + y^2 &= 1 \end{align}

I am having difficulties solving the system of equations. I got $\lambda = 1$ from $2y = \lambda 2y$ and from there I obtain $x = \frac{1}{4}$. However I don't understand how I am supposed to solve for a value of $y$.

I tried to plug in $x = 1/4$ into $2x^2 + 2y^2 = 1$ but I got $y = \sqrt{7/4}$ which does not fit the system of equations. I think my setup is correct and I am doing the calculations wrong. I think I should be getting $x = 1/4$ and $y = \sqrt{7}/4$. What do you guys think?

Answer 1

From the equality $2y=2\lambda y$, you can deduce that either $y=0$ or that $\lambda=1$.

If $\lambda=1$, your first and third equations become $4x=1$ and $2x^2+y^2=1$. In this case, the solutions are $(x,y)=\left(\frac14,\pm\sqrt{\frac78}\right)$.

If $y=0$, your first and third equations become $4\lambda x=1$ and $2x^2=1$. In this case, the solutions are $(x,y)=\left(\pm\frac1{\sqrt2},0\right)$.

Answer 2

Did you plug $x=\frac14$ in $2x^2+2y^2=1$ instead of $2x^2+y^2=1$?

Answer 3

Actually you do not need Lagrange multipliers: by setting $x=\frac{1}{\sqrt{2}}z$ you are looking for the minimum of $\frac{1}{\sqrt{2}}z+y^2$ under the constraint $z^2+y^2=1$, so you are looking for the minimum of $\frac{1}{\sqrt{2}}z+(1-z^2)$ over $[-1,1]$, which is achieved at the left endpoint of the domain.