Using the second partial derivative test to find extrema in D :
Compute the partial derivatives of f(x, y) = 2x² + y⁴.
∂f/∂x = 4x
∂f/∂y = 4y³
Find the critical points of f, where both partial derivatives vanish.
4x = 0 ⇒ x = 0
4y³ = 0 ⇒ y = 0
So f has only one critical point at (0, 0), which does belong to the set D.
Compute the determinant of the Hessian matrix of f at (0, 0) :
[tex]H = \begin{bmatrix}\dfrac{\partial^2f}{\partial x^2} & \dfrac{\partial^2f}{\partial y\partial x} \\ \\ \dfrac{\partial^2f}{\partial x\partial y} & \dfrac{\partial^2f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}4 & 0 \\ 0 & 12y^2 \end{bmatrix}[/tex]
We have det(H) = 48y² = 0 at the origin, which means the second partial derivative test fails. However, we observe that 2x² + y⁴ ≥ 0 for all x, y because the square of any real number cannot be negative, so (0, 0) must be a minimum and we have f(0, 0) = 0.
Using the second derivative test to find extrema on the boundary of D :
Let x = cos(t) and y = sin(t) with 0 ≤ t < 2π, so that (x, y) is a point on the circle x² + y² = 1. Then
f(cos(t), sin(t)) = g(t) = 2 cos²(t) + sin⁴(t)
is a function of a single variable t. Find its critical points, where the first derivative vanishes.
g'(t) = -4 cos(t) sin(t) + 4 sin³(t) cos(t) = 0
⇒ cos(t) sin(t) (1 - sin²(t)) = 0
⇒ cos³(t) sin(t) = 0
⇒ cos³(t) = 0 or sin(t) = 0
⇒ cos(t) = 0 or sin(t) = 0
⇒ [t = π/2 or t = 3π/2] or [t = 0 or t = π]
Check the values of g'' at each of these critical points. We can rewrite
g'(t) = -4 cos³(t) sin(t)
Then differentiating yields
g''(t) = 12 cos²(t) sin²(t) - 4 cos⁴(t)
g''(0) = 12 cos²(0) sin²(0) - 4 cos⁴(0) = -4
g''(π/2) = 12 cos²(π/2) sin²(π/2) - 4 cos⁴(π/2) = 0
g''(π) = 12 cos²(π) sin²(π) - 4 cos⁴(π) = -4
g''(3π/2) = 12 cos²(3π/2) sin²(3π/2) - 4 cos⁴(3π/2) = 0
Since g''(0) and g''(π) are both negative, the points (x, y) corresponding to t = 0 and t = π are maxima.
t = 0 ⇒ x = cos(0) = 1 and y = sin(0) = 0 ⇒ f(1, 0) = 2
t = π ⇒ x = cos(π) = -1 and y = sin(π) = 0 ⇒ f(-1, 0) = 2
Both g''(π/2) and g''(3π/2) are zero, so the test fails. These values of t correspond to
t = π/2 ⇒ x = cos(π/2) = 0 and y = sin(π/2) = 1 ⇒ f(0, 1) = 1
t = 3π/2 ⇒ x = cos(3π/2) = 0 and y = sin(3π/2) = -1 ⇒ f(0, -1) = 1
but both of the values of f at these points are between the minimum we found at 0 and the maximum at 2.
So over the region D, max(f) = 2 at (±1, 0) and min(f) = 0 at (0, 0).
