The problem asks to find the global maximum and minimum of the function $f(x, y) = x^2 + y^2 - 2x - 4y$ on the triangular region $D$ defined by $D = \{ (x, y) \mid x \geq 0, y \geq 0, x + y \leq 3 \}$.

Steps to solve the problem:

1. Find the critical points within the region $D$:

   The critical points are found by setting the gradient of $f$ equal to zero:

   $\nabla f(x, y) = (2x - 2, 2y - 4)$

   Setting the gradient equal to zero:

   $2x - 2 = 0 \implies x = 1, \quad 2y - 4 = 0 \implies y = 2$

   So, the critical point is $(1, 2)$.

2. Check if the critical point is inside the region $D$:

   Since $1 \geq 0$, $2 \geq 0$, and $1 + 2 = 3$, the critical point $(1, 2)$ lies inside the region $D$.

3. Evaluate the function at the critical point:

   $f(1, 2) = 1^2 + 2^2 - 2 \cdot 1 - 4 \cdot 2 = 1 + 4 - 2 - 8 = -5$

4. Evaluate the function on the boundary of the region $D$:

   • Boundary 1: $x = 0$ (from $y = 0$ to $y = 3$)

     $f(0, y) = 0^2 + y^2 - 2 \cdot 0 - 4y = y^2 - 4y$

     To find the minimum and maximum, take the derivative and set to zero:

     $\frac{d}{dy}(y^2 - 4y) = 2y - 4 = 0 \implies y = 2$

     • $f(0, 0) = 0$
     • $f(0, 2) = -4$
     • $f(0, 3) = -3$

   • Boundary 2: $y = 0$ (from $x = 0$ to $x = 3$)

     $f(x, 0) = x^2 - 2x$

     To find the minimum and maximum, take the derivative and set to zero:

     $\frac{d}{dx}(x^2 - 2x) = 2x - 2 = 0 \implies x = 1$

     • $f(0, 0) = 0$
     • $f(1, 0) = -1$
     • $f(3, 0) = 3$

   • Boundary 3: $x + y = 3$ (with $x \geq 0, y \geq 0$)

     Let $y = 3 - x$:

     $f(x, 3 - x) = x^2 + (3 - x)^2 - 2x - 4(3 - x)$

     Simplify:

     $f(x, 3 - x) = x^2 + 9 - 6x + x^2 - 2x - 12 + 4x = 2x^2 - 4x - 3$

     To find the minimum and maximum, take the derivative and set to zero:

     $\frac{d}{dx}(2x^2 - 4x - 3) = 4x - 4 = 0 \implies x = 1$

     • $f(0, 3) = -3$
     • $f(1, 2) = -5$
     • $f(3, 0) = 3$

Conclusion:

• The global minimum value is $-5$ at the point $(1, 2)$.
• The global maximum value is $3$ at the point $(3, 0)$.

Would you like more details or have any questions about this solution?

Related Questions:

1. How do you find critical points of a function within a region?
2. What is the significance of evaluating functions on boundaries of a region?
3. How does one apply the method of Lagrange multipliers in optimization problems?
4. What is the difference between local and global extrema?
5. How can optimization be extended to higher dimensions?

Tip: Always check the boundaries of the region when finding extrema in a constrained problem, as maximum or minimum values often occur there.