We first compute the partial derivatives of f(x, y):
f(x, y) = 5x² + 5y² - 10x + 21
∂f/∂x = 10x - 10
∂f/∂y = 10y
The critical points of f occur where both ∂f/∂x and ∂f/∂y are equal to zero. This only happens at the point (x, y) = (1, 0). (And this point does lie inside R.)
Next, compute the Hessian matrix H(x, y) for f :
[tex]H(x, y) = \begin{bmatrix}\frac{\partial^2f}{\partial x^2} & \frac{\partial^2f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}10&0\\0&10\end{bmatrix}[/tex]
Since det(H(x,y)) = 100 is positive for all x and y, this means the critical point (1, 0) is a local minimum, and we have f(1, 0) = 16.
Next, we check for extrema along the boundary of R, which is comprised of a semicircle with radius 2 and the line segment connecting (-2, 0) and (2, 0).
• Parameterize the semicircular portion by
x(t) = 2 cos(t)
y(t) = 2 sin(t)
with 0 ≤ t ≤ π. Then
f(x(t), y(t)) = 20 sin²(t) + 20 cos²(t) - 20 cos(t) + 21
which is simplifies to a function of one variable t,
g(t) = 41 - 20 cos(t)
Find the extrema of g over the interval [0, π] : we have critical points when
g'(t) = 20 sin(t) = 0
which happens at t = 0 and t = π. At these points, we get local extreme values of
•• t = 0 => x = 2 and y = 0 => f(2, 0) = 21
•• t = π => x = -2 and y = 0 => f(-2, 0) = 61
• Over the line-segment portion, we take y = 0, so f(x, y) again reduces to a function of one variable:
f(x, 0) = 5x² - 10x + 21
Completing the square, we have
5x² - 10x + 21 = 5 (x - 1)² + 16
which has a maximum value of 16 when x = 1 (and this happens to be the critical point (1, 0) we found earlier).
So, over the region R, f(x, y) has
• an absolute maximum of 61 at the point (-2, 0), and
• an absolute minimum of 16 at (1, 0)
