Determining constrained extrema

Problem 5, Section 14.8, Stewart's Calculus, 8e

Find the extreme values of the function $f (x, y) = x y$ subject to the constraint $4 x^{2} + y^{2} = 8$ .

Let $g (x, y) = 4 x^{2} + y^{2}$ . The constrained critical points are those points that satisfy $g (x, y) = 8$ and at which the gradient vectors $\nabla f$ and $\nabla g$ are parallel. So we first compute those gradient vectors:

\nabla f = [\begin{matrix} y \\ x \end{matrix}] and \nabla g = [\begin{matrix} 8 x \\ 2 y \end{matrix}] .

Using the "determinant trick", these vectors are parallel exactly when the determinant of the $2 \times 2$ matrix with these vectors as its column vanishes. So we compute

det [\begin{matrix} y & 8 x \\ x & 2 y \end{matrix}] = y \cdot 2 y - 8 x \cdot x = 2 y^{2} - 8 x^{2} .

So now we need to solve the system of equations

\begin{aligned} 4 x^{2} + y^{2} & = 8 \\ 2 y^{2} - 8 x^{2} & = 0. \end{aligned}

There are various ways we can solve these two equations. For instance, we might solve the second equation for $y^{2}$ to get $y^{2} = 4 x^{2}$ , and then substitute that into the first equation to get $4 x^{2} + (4 x^{2}) = 8$ , which simplifies to $x^{2} = 1$ , and so either $x = 1$ or $x = - 1$ . Since $y^{2} = 4 x^{2}$ , both of these cases imply $y^{2} = 4$ , and so either $y = 2$ or $y = - 2$ . Note that this means we've found four points, namely $(1, 2), (1, - 2), (- 1, 2)$ , and $(- 1, - 2)$ .

To determine the max and min values of $f$ , we now simply evaluate $f$ at each of these four points. The highest value is the max and the lowest value is the min.