844b65…: “Aboard the windy flower”

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Suppose that $F(x,y) =x^2-xy+y^2+4y$ . Then, $\grad {f}(x,y) = \vector {\answer {2x-y},\answer {2y-x+4}}.$

How many distinct points $(x,y)$ are there for which $\grad {F}(x,y) = \vec {0}$ ?

none one two more than two, but finitely many infinitely many

If $\grad {F}(x,y) = \vec {0}$ , we must have that the $x$ and $y$ components are zero simultaneously. We thus must have that $\begin{align*} 2x-y&=0 \\ 2y-x+4 &=0 \end{align*}$

How many options are there?

The point $(x,y)$ for which $\grad {F}(x,y) = \vec {0}$ is $(x,y) = \left (\answer {-\frac {4}{3}},\answer {-\frac {8}{3}} \right )$

We must solve the system of equations $\begin{align} 2x-y&=0 \\ x-2y &=4 \end{align}$

This can be done many ways. Multiplying the top equation by $2$ and adding the second equation to it gives $\answer {3} \cdot x = \answer {-4}$ , from which we find $x = \answer {-\frac {4}{3}}$ . We can now use either equation to find $y$ .

As it turns out, the points for which $\grad {F}(x,y) = \vec {0}$ will be very important when we try to find relative extrema of a function of two variables. This exercise gives a little practice with analyzing the gradient to that end.