Question Two: a. Define the following: i. Local/Relative Maximum ii. Local/Relative Minimum iii. Critical Point. b. Find and classify the relative extrema for the function F(x,y)=-x2+4y-2y3+1. c. Find the absolute maximum/minimum of f(x,y)=4x+6y−x2−y2 on the boundary B = [(x,y)|0<=x<=4,0<=y<=5].

$$F_x = \frac{\partial}{\partial x}(-x^2 + 4xy - 2y^2 + 1) = -2x + 4y$$ $$F_y = \frac{\partial}{\partial y}(-x^2 + 4xy - 2y^2 + 1) = 4x - 4y$$

$$-2x + 4y = 0 \implies x = 2y \quad (1)$$ $$4x - 4y = 0 \implies x = y \quad (2)$$

Substitute (2) into (1):

$$y = 2y \implies y = 0$$

Then $x = 0$.
The only critical point is $(0,0)$.

First, find the second partial derivatives:

$$F_{xx} = \frac{\partial}{\partial x}(-2x + 4y) = -2$$ $$F_{yy} = \frac{\partial}{\partial y}(4x - 4y) = -4$$ $$F_{xy} = \frac{\partial}{\partial y}(-2x + 4y) = 4$$

Calculate the discriminant $D = F_{xx}F_{yy} - (F_{xy})^2$:

$$D = (-2)(-4) - (4)^2 = 8 - 16 = -8$$

Since $D < 0$, the critical point $(0,0)$ is a saddle point.
Therefore, there are no relative maxima or minima for the function $F(x,y)$.

The relative extrema are: $\boxed{\text{None}}$

$$f_x = \frac{\partial}{\partial x}(4x + 6y - x^2 - y^2) = 4 - 2x$$ $$f_y = \frac{\partial}{\partial y}(4x + 6y - x^2 - y^2) = 6 - 2y$$

Set $f_x = 0 \implies 4 - 2x = 0 \implies x = 2$.
Set $f_y = 0 \implies 6 - 2y = 0 \implies y = 3$.
The critical point is $(2,3)$. This point lies within the given region $0 \le x \le 4, 0 \le y \le 5$.

The region is a rectangle with corners $(0,0), (4,0), (0,5), (4,5)$.

• Boundary 1: $x=0$, $0 \le y \le 5$ $f(0,y) = 6y - y^2$. Let $g(y) = 6y - y^2$. $g'(y) = 6 - 2y = 0 \implies y = 3$. Points to check: $(0,3)$, and the endpoints $(0,0), (0,5)$. $f(0,3) = 6(3) - (3)^2 = 18 - 9 = 9$. $f(0,0) = 0$. $f(0,5) = 6(5) - (5)^2 = 30 - 25 = 5$.

• Boundary 2: $x=4$, $0 \le y \le 5$ $f(4,y) = 4(4) + 6y - (4)^2 - y^2 = 16 + 6y - 16 - y^2 = 6y - y^2$. This is the same as Boundary 1. Critical point at $y=3$. Points to check: $(4,3)$, and the endpoints $(4,0), (4,5)$. $f(4,3) = 6(3) - (3)^2 = 18 - 9 = 9$. $f(4,0) = 0$. $f(4,5) = 6(5) - (5)^2 = 30 - 25 = 5$.

• Boundary 3: $y=0$, $0 \le x \le 4$ $f(x,0) = 4x - x^2$. Let $h(x) = 4x - x^2$. $h'(x) = 4 - 2x = 0 \implies x = 2$. Points to check: $(2,0)$, and the endpoints $(0,0), (4,0)$. $f(2,0) = 4(2) - (2)^2 = 8 - 4 = 4$. $f(0,0) = 0$ (already found). $f(4,0) = 0$ (already found).

• Boundary 4: $y=5$, $0 \le x \le 4$ $f(x,5) = 4x + 6(5) - x^2 - (5)^2 = 4x + 30 - x^2 - 25 = 4x - x^2 + 5$. Let $k(x) = 4x - x^2 + 5$. $k'(x) = 4 - 2x = 0 \implies x = 2$. Points to check: $(2,5)$, and the endpoints $(0,5), (4,5)$. $f(2,5) = 4(2) - (2)^2 + 5 = 8 - 4 + 5 = 9$. $f(0,5) = 5$ (already found). $f(4,5) = 5$ (already found).

The values of $f(x,y)$ at the critical point and boundary points are:
$f(2,3) = 13$
$f(0,3) = 9$
$f(0,0) = 0$
$f(0,5) = 5$
$f(4,3) = 9$
$f(4,0) = 0$
$f(4,5) = 5$
$f(2,0) = 4$
$f(2,5) = 9$

The maximum value is 13.
The minimum value is 0.

The absolute maximum is $\boxed{13}$ and the absolute minimum is $\boxed{0}$.