Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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).
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
