Question

Find the absolute maximum and minimum values of function f on the set D.f(x,y) = e^-x^2-y^2()x^2+2y^2; D is the disk x^2+y^2<=4

Answer 1

You probably mean the function to be

$f(x, y) = e^{-x^2-y^2} (x^2 + 2y^2)$

We use the second (partial) derivative test for this.

Compute the first partial derivatives of f :

$\dfrac{\partial f}{\partial x} = e^{-x^2-y^2}(2x - 2x(x^2+2y^2)) = e^{-x^2-y^2} (2x - 2x^3 + 4xy^2)$

$\dfrac{\partial f}{\partial y} = e^{-x^2-y^2}(4y - 2y(x^2+2y^2)) = e^{-x^2-y^2} (4y-4y^3-2x^2y)$

Find the critical points of f, where both first derivatives vanish. Since $e^{-x^2-y^2}\ge0$ for all x and y, we're left with

$2x - 2x^3 - 4xy^2 = 2x (1 - x^2 - 2y^2)= 0$

which means either x = 0 or x² + 2y² = 1, and

$4y - 4y^3 - 2x^2y = 2y (2 - 2y^2 - x^2) = 0$

so y = 0 or x² + 2y² = 2.

• If x = 0 and y = 0, then one critical point is (0, 0).

• If x ≠ 0 and y = 0, then x² + 2y² = x² = 1, which means x = ±1. So we have two additional critical points, (-1, 0) and (0, 0).

• If x = 0 and y ≠ 0, then x² + 2y² = 2y² = 2, so y² = 1 and y = ±1, and we get two more critical points at (0, -1) and (0, 1).

All five of these critical points belong to D.

Compute the Hessian matrix for f and check the sign of its determinant at each critical point. Remember what the second partial derivative test concludes:

• If det H(x, y) is negative, then that point is a saddle point

• If det H(x, y) is positive, and ∂²f/∂x² is negative, then that point is a maximum

• If det H(x, y) is positive, and ∂²f/∂x² is positive, then that point is a minimum

The Hessian matrix for f is

$H(x,y) = \begin{bmatrix}\frac{\partial^2f}{\partial x^2} & \frac{\partial^2f}{\partial y\partial x} \\ \frac{\partial^2f}{\partial x\partial y} & \frac{\partial^2f}{\partial y^2}\end{bmatrix}$

$H(x,y) = \begin{bmatrix}2 - 10 x^2 + 4 x^4 - 4 y^2 + 8 x^2 y^2 & -12 x y + 4 x^3 y + 8 x y^3 \\ -12 x y + 4 x^3 y + 8 x y^3 & 4 - 2 x^2 - 20 y^2 + 4 x^2 y^2 + 8 y^4\end{bmatrix} e^{-x^2-y^2}$

Note that ∂²f/∂x² is the (1, 1) entry. Compute the determinant and evaluate it at each critical point.

$\det H(0,0) = \det\begin{bmatrix}2&0\\0&4\end{bmatrix} = 8 > 0$

$\dfrac{\partial^2f}{\partial x^2} (0, 0) = 2 > 0$

⇒   local minimum at (0, 0) of f(0, 0) = 0

$\det H(-1, 0) = \det\begin{bmatrix}-\frac4e & 0 \\ 0 & \frac2e\end{bmatrix} = -\dfrac8{e^2} < 0$

⇒   saddle point (-1, 0)

$\det H(1, 0) = \det\begin{bmatrix}-\frac4e & 0 \\ 0 & \frac2e\end{bmatrix} = -\dfrac8{e^2} < 0$

⇒   saddle point at (1, 0)

$\det H(0,-1) = \det\begin{bmatrix}-\frac2e & 0 \\ 0 & -\frac8e\end{bmatrix} = \dfrac{16}{e^2} > 0$

$\dfrac{\partial^2f}{\partial x^2} (0,-1) = -\dfrac2e < 0$

⇒   local maximum at (0, -1) of f(0, -1) = 2/e ≈ 0.736

$\det H(0,1) = \det\begin{bmatrix}-\frac2e & 0 \\ 0 & -\frac8e\end{bmatrix} = \dfrac{16}{e^2} > 0$

$\dfrac{\partial^2f}{\partial x^2} (0,1) = -\dfrac2e < 0$

⇒   local maximum at (0, 1) of f(0, 1) = 2/e ≈ 0.736

Check for extrema along the boundary. We can parameterize the boundary by setting

x(t) = 2 cos(t)

y(t) = 2 sin(t)

with 0 ≤ t < 2π. Then f(x, y) = f(x(t), y(t)) reduces to a function of only t, which we call F(t) :

$f(2 \cos(t), 2 \sin(t)) = e^{-4\cos^2(t)-4\sin^2(t)} (4\cos^2(t) + 8 \sin^2(t))$

$\implies F(t) = \dfrac4{e^4} (1 + \sin^2(t)) = \dfrac2{e^4} (3 - \cos(2t))$

Find the critical points of F on the interval [0, 2π) :

$F'(t) = \dfrac4{e^4} \sin(2t) = 0$

sin(2t) = 0

2t = arcsin(0) + 2nπ   or   2t = π - arcsin(0) + 2nπ

(where n is any integer)

2t = 2nπ   or   2t = π + 2nπ

t = nπ   or   t = π/2 + nπ

Over [0, 2π), critical points occur at t = 0, t = π/2, t = π, and t = 3π/2.

Check the sign of the second derivative of F at each critical point. The test in the one-variable case says

• if F''(t) < 0, then that point is a maximum

• if F''(t) > 0, then that point is a minimum

$F''(t) = \dfrac8{e^4} \cos(2t)$

$F''(0) = \dfrac8{e^4} > 0$

⇒   local minimum when t = 0 of F(0) = 4/e⁴ ≈ 0.0733. This t corresponds to the point (2, 0), since x(0) = 2 and y(0) = 0.

$F''\left(\dfrac\pi2\right) = -\dfrac8{e^4} < 0$

⇒   local maximum when t = π/2 of F(π/2) = 8/e⁴ ≈ 0.147, referring to the point (0, 2)

$F''(\pi) = \dfrac8{e^4} > 0$

⇒   local minimum when t = π of F(π) = 4/e⁴ ≈ 0.0733 at the point (-2, 0)

$F''\left(\dfrac{3\pi}2\right) = -\dfrac8{e^4} < 0$

⇒   local maximum when t = 3π/2 of F(3π/2) = 8/e⁴ ≈ 0.147 at (0, -2)

So, the absolute extrema of f(x, y) over D are

• minimum : f(0, 0) = 0

• maximum : f(0, -1) = f(0, 1) = 2/e ≈ 0.736