Find the absolute maximum and minimum values of the following function on the specified region R. F(x,y) = 9xy on the semicircular disk.
R = {(x, y): -2 ≤ x ≤ 2, 0 ≤ y ≤ (4 - x²):}

Solution:

Given, F(x, y) = 9xy on the semicircular disk.

R = {(x, y): -2 ≤ x ≤ 2, 0 ≤ y ≤ (4 - x²):}

We have to find the absolute maximum and minimum values of the function.

Since the region R is a semicircular disc, we assume the boundaries of the region are given by -2 ≤ x ≤ 2, 0 ≤ y ≤ √(4 - x²)

To find the point for which the gradient is equal to zero,

\(\frac{dF}{dx}=9y=0\\\frac{dF}{dy}=9x=0\)

This implies that (x, y) = (0, 0) lies inside the region R.

Using Hessian criteria to check maximum or minimum, we calculate the matrix of second derivatives

\(\frac{d^{2}F}{dx^{2}}=0=\frac{d^{2}F}{dy^{2}}\)

\(\frac{d^{2}F}{dx dy}=9=\frac{d^{2}F}{dy dx}\)

We get the matrix \(\bigl(\begin{smallmatrix} 0 &9 \\ 9 &0 \end{smallmatrix}\bigr)\)

The first determinant is zero and the second determinant is -9.

Thus, the point is a saddle point.

Hence, the point is not a minimum or maximum.

Since the function is continuous and the region R is closed and bound the maximum and minimum must be attained on the boundaries of R.

When -2 ≤ x ≤ x and y = 0, we have that F(x, 0) = 0. We want to pay attention to the critical values over the circle, restricting that the values of y must be positive.

Consider H(x, y, λ) = 9xy - λ(x² + y²- 4) that describes the restriction (x² + y²= 4), we want that the gradient of H is 0.

Then, \(\frac{dF}{dx}=9y-2\lambda x=0\)

\(\frac{dF}{dx}=9x-2\lambda y=0\)

\(\frac{dF}{d\lambda }=x^{2}+y^{2}-4=0\)

From the first and second equation we get that

\(\lambda =\frac{9y}{2x}=\frac{9x}{2y}\)

Which implies that y² = x².

Put the value of y² in the restriction we get,

x² + x² = 2x² = 4

x² = 4/2

x² = 2

Taking square root,

x = ±√2

Since y is only positive, the critical points are (√2, √2) and (-√2, √2)

For the first point, the value of the function is

F(√2, √2) = 9(2) = 18

For the second point, the value of the function is

F(-√2, √2) = 9(-2) = -18

Therefore, the maximum point is (√2, √2) and minimum point is (-√2, √2).