938f36…: “Except a foggy water”

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Use Lagrange multipliers to find the extrema of $F(x,y) = 3y-2x^2$ constrained the parabola $y=x^2+x-1$ . Start by setting $G(x,y) = y-x^2-x+1$ . Now compute: $\begin{align*} \grad F(x,y) &=\vector{\answer{-4x},\answer{3}}\\ \grad G(x,y) &=\vector{\answer{-2x-1},\answer{1}} \end{align*}$

Now solve the following system of equations for $x$ , $y$ , and $\lambda$ : $\begin{align*} \grad F(x,y) &= \lambda \cdot \grad G(x,y)\\ y&= x^2+x-1 \end{align*}$

The solutions will be the extrema on the parabola. $\vec {e} = \vector {\answer {-3/2},\answer {-1/4}}$ This point corresponds to a:

Minimum Maximum

The value of $F$ at this point is $\answer {-21/4}$ .