Find the maximum number of bicycle of each type to make maximum profit

Let $x$ be the number of type A bicycles and $y$ be the number of type B bicycles.

The profit for a type A bicycle is K10 000, and for a type B bicycle is K15 000.
The objective function to maximize profit $P$ is:

$$P = 10000x + 15000y$$

From the previous part, the constraints are:

1. Cost constraint: $3x + 4y \le 120$
2. Storage constraint: $x + y \le 40$
3. Non-negativity: $x \ge 0$
4. Non-negativity: $y \ge 0$

The feasible region is defined by the intersection of these inequalities. We find the corner points (vertices) of this region.

• Intersection of $x=0$ and $y=0$: $(0,0)$

• Intersection of $y=0$ and $3x + 4y = 120$: $3x + 4(0) = 120 \implies 3x = 120 \implies x = 40$. Vertex: $(40,0)$

• Intersection of $x=0$ and $3x + 4y = 120$: $3(0) + 4y = 120 \implies 4y = 120 \implies y = 30$. Vertex: $(0,30)$

• Intersection of $x + y = 40$ and $3x + 4y = 120$: From $x + y = 40$, we can write $x = 40 - y$. Substitute this into the second equation: $3(40 - y) + 4y = 120$ $120 - 3y + 4y = 120$ $120 + y = 120$ $y = 0$ Substitute $y=0$ back into $x = 40 - y$: $x = 40 - 0 = 40$. Vertex: $(40,0)$

The constraint $x+y \le 40$ is redundant because the feasible region is already bounded by $3x+