Algebra - Linear programming, including inequalities-extended

A small factory produces two products, $x$ and $y$. The constraints are given by: $x + y \leq 10$, $x \geq 2$, and $y \geq 3$. Find the feasible region and maximize the profit function $P = 4x + 5y$.

First, identify the vertices of the feasible region by finding the intersection points of the boundary lines:

1. Intersection of $x = 2$ and $y = 3$: Point $A(2, 3)$
2. Intersection of $x = 2$ and $x + y = 10$: $2 + y = 10 \implies y = 8$. Point $B(2, 8)$
3. Intersection of $y = 3$ and $x + y = 10$: $x + 3 = 10 \implies x = 7$. Point $C(7, 3)$

Now, evaluate $P = 4x + 5y$ at each vertex:

• At $A(2, 3)$: $P = 4(2) + 5(3) = 8 + 15 = 23$
• At $B(2, 8)$: $P = 4(2) + 5(8) = 8 + 40 = 48$
• At $C(7, 3)$: $P = 4(7) + 5(3) = 28 + 15 = 43$

The maximum profit is 48 at point $(2, 8)$.

We first convert the inequalities into boundary equations to find the corners of the shape formed by the constraints. Since the maximum must occur at a corner, we plug those coordinates into the profit formula and compare the results.

Show the region satisfied by the inequality $2x - 3y > 6$.

1. Find the boundary line by treating it as an equation: $2x - 3y = 6$.
2. Find intercepts:
   • If $x = 0$, $-3y = 6 \implies y = -2$. Point: $(0, -2)$
   • If $y = 0$, $2x = 6 \implies x = 3$. Point: $(3, 0)$
3. Draw a dashed line through $(0, -2)$ and $(3, 0)$ because the inequality is strict ($>$).
4. Test a point not on the line, e.g., $(0, 0)$: $2(0) - 3(0) > 6 \implies 0 > 6$ This is false, so shade the region that does not contain $(0, 0)$.

The boundary line splits the plane into two halves. A test point helps determine which side satisfies the inequality. A dashed line is essential for '$>$' because points on the line are not included in the solution set.