A school supplier has 21000/=. He wants to buy Kiswahili and Business text books. Kiswahili text books cost 280/= each and Business text books cost 350/= each. He has to buy at least 26 Kiswahili text

To represent the inequalities on a grid, we first convert them to equations to find the boundary lines.

1. For $4x + 5y \le 300$: Equation: $4x + 5y = 300$ If $x = 0$, $5y = 300 \implies y = 60$. Point: $(0, 60)$ If $y = 0$, $4x = 300 \implies x = 75$. Point: $(75, 0)$ Test point $(0,0)$: $4(0) + 5(0) = 0 \le 300$. So, shade the region towards the origin.

2. For $x \ge 26$: Equation: $x = 26$ This is a vertical line passing through $x=26$. Shade the region to the right of this line.

3. For $x < 2y$ (or $2y > x$): Equation: $x = 2y$ (or $y = \frac{1}{2}x$) If $x = 0$, $y = 0$. Point: $(0, 0)$ If $x = 50$, $y = 25$. Point: $(50, 25)$ If $x = 60$, $y = 30$. Point: $(60, 30)$ This is a dashed line because of the strict inequality ($<$). Test point $(20, 0)$: $20 < 2(0) \implies 20 < 0$, which is false. So, shade the region above the line (away from $(20,0)$).

The feasible region will be the area where all shaded regions overlap.
(A graph would be drawn here. The x-axis represents Kiswahili books ($x$) and the y-axis represents Business books ($y$). The scale should accommodate $x$ up to 75 and $y$ up to 60. The feasible region will be a polygon bounded by these lines.)

The question asks for minimum cost, but the objective function for cost is $C = 280x + 350y$. The constraint $4x + 5y \le 300$ already represents the maximum budget. If we want to minimize cost, we would ideally buy 0 books, but the problem states "He has to buy at least 26 Kiswahili text books" and implies buying books. This usually means minimizing cost while satisfying all conditions and possibly a minimum total number of books, or maximizing profit/minimizing cost within the feasible region. Given the phrasing "minimum cost", it implies we are looking for the point in the feasible region that results in the lowest cost.

Let's re-read the question carefully: "Determine the number of text book of each kind that must be bought for the total must be minimum cost." This implies we are looking for the minimum cost within the feasible region defined by the inequalities.

The vertices of the feasible region are the points where the boundary lines intersect. We need to find these vertices and evaluate the cost function $C = 280x + 350y$ at each vertex.

Let's find the intersection points:

Intersection 1: $x = 26$ and $4x + 5y = 300$
Substitute $x = 26$ into $4x + 5y = 300$:
$4(26) + 5y = 300$
$104 + 5y = 300$
$5y = 300 - 104$
$5y = 196$
$y = \frac{196}{5} = 39.2$
Since the number of books must be an integer, we consider points near $(26, 39.2)$. For $x=26$, $y$ must be an integer.
If $y=39$, $4(26) + 5(39) = 104 + 195 = 299 \le 300$. This point $(26, 39)$ is in the feasible region.
If $y=40$, $4(26) + 5(40) = 104 + 200 = 304 > 300$. This point is outside.
So, the point is $(26, 39)$.

Intersection 2: $x = 26$ and $x = 2y$
Substitute $x = 26$ into $x = 2y$:
$26 = 2y$
$y = 13$
Point: $(26, 13)$

Intersection 3: $4x + 5y = 300$ and $x = 2y$
Substitute $x = 2y$ into $4x + 5y = 300$:
$4(2y) + 5y = 300$
$8y + 5y = 300$
$13y = 300$
$y = \frac{300}{13} \approx 23.07$
$x = 2y = 2 \times \frac{300}{13} = \frac{600}{13} \approx 46.15$
Since the number of books must be an integer, we consider points near $(\frac{600}{13}, \frac{300}{13})$.
Let's check integer points around this intersection.
If $y=23$, $x=2y=46$.
Check $4x+5y \le 300$: $4(46) + 5(23) = 184 + 115 = 299 \le 300$. This point $(46, 23)$ is in the feasible region.
Check $x < 2y$: $46 < 2(23) \implies 46 < 46$, which is false. So $(46, 23)$ is on the line $x=2y$, not strictly less than.
We need $x < 2y$. So we need to consider points where $x$ is slightly less than $2y$.
Let's try $x=45, y=23$.
Check $x < 2y$: $45 < 2(23) \implies 45 < 46$. This is true.
Check $x \ge 26$: $45 \ge 26$. This is true.
Check $4x+5y \le 300$: $4(45) + 5(23) = 180 + 115 = 295 \le 300$. This is true.
So, $(45, 23)$ is a feasible point.

The vertices of the feasible region are approximately:
A: Intersection of $x=26$ and $x=2y \implies (26, 13)$
B: Intersection of $x=26$ and $4x+5y=300 \implies (26, 39.2)$. Since $y$ must be an integer, we use $(26, 39)$.
C: Intersection of $4x+5y=300$ and $x=2y$. This is tricky due to the strict inequality $x < 2y$. The line $x=2y$ forms the boundary, but points on the line are not included.
The feasible region is a polygon. The vertices are the points where the boundary lines intersect.
The vertices are:

1. $(26, 13)$ (from $x=26$ and $x=2y$)
2. $(26, 39)$ (from $x=26$ and $4x+5y=300$, taking integer $y$)
3. The third vertex would be where $4x+5y=300$ and $x=2y$ intersect, which is $(\frac{600}{13}, \frac{300}{13})$. However, since $x < 2y$ is a strict inequality, this point itself is not in the feasible region. We need to consider integer points just inside the feasible region.

Let's list the integer vertices of the feasible region.
The feasible region is bounded by $x=26$, $y=\frac{1}{2}x$, and $y=\frac{300-4x}{5}$.
The vertices are:

1. Point A: Intersection of $x=26$ and $y=\frac{1}{2}x$. $x=26 \implies y = \frac{1}{2}(26) = 13$. So, $(26, 13)$. Check $4x+5y \le 300$: $4(26) + 5(13) = 104 + 65 = 169 \le 300$. This is a valid vertex.

2. Point B: Intersection of $x=26$ and $y=\frac{300-4x}{5}$. $x=26 \implies y = \frac{300-4(26)}{5} = \frac{300-104}{5} = \frac{196}{5} = 39.2$. Since $y$ must be an integer, we consider the point $(26, 39)$. Check $x < 2y$: $26 < 2(39) \implies 26 < 78$. This is true. So, $(26, 39)$ is a valid vertex.

3. Point C: Intersection of $y=\frac{1}{2}x$ and $y=\frac{300-4x}{5}$. $\frac{1}{2}x = \frac{300-4x}{5}$ $5x = 2(300-4x)$ $5x = 600 - 8x$ $13x = 600$ $x = \frac{600}{13} \approx 46.15$ $y = \frac{1}{2}x = \frac{300}{13} \approx 23.07$ Since $x < 2y$ is a strict inequality, the point $(\frac{600}{13}, \frac{300}{13})$ is not included. We need to find integer points in the feasible region near this intersection. Let's consider integer points $(x,y)$ such that $x < 2y$ and $4x+5y \le 300$. If $x=46$, then $y$ must be at least 23.5 for $x < 2y$. So $y=24$. Check $(46, 24)$: $x \ge 26$: $46 \ge 26$. True. $x < 2y$: $46 < 2(24) \implies 46 < 48$. True. $4x+5y \le 300$: $4(46) + 5(24) = 184 + 120 = 304$. This is $> 300$, so $(46, 24)$ is outside the budget constraint. This means the feasible region is bounded by $x=26$, $y=\frac{1}{2}x$ and $4x+5y=300$. The "vertex" where $4x+5y=300$ and $x=2y$ meet is not an integer point and is not strictly in the region. The feasible region is a polygon with vertices at $(26,13)$, $(26,39)$, and points along the line $4x+5y=300$ and $x=2y$ that satisfy the integer constraint and strict inequality.

Let's re-evaluate the vertices for integer solutions.
The feasible region is defined by:

1. $x \ge 26$
2. $y > \frac{1}{2}x$
3. $y \le \frac{300-4x}{5}$

We need to find integer points $(x,y)$ that satisfy these conditions.
The vertices of the feasible region are the points where the boundary lines intersect.
Vertex 1: Intersection of $x=26$ and $y=\frac{1}{2}x$.
$x=26 \implies y = \frac{1}{2}(26) = 13$. Point $(26, 13)$.
This point satisfies $x \ge 26$ and $y = \frac{1}{2}x$. However, the inequality is $y > \frac{1}{2}x$. So $(26, 13)$ is not strictly in the feasible region. We need to consider points like $(26, 14)$.
Let's check $(26, 14)$:
$x \ge 26$: $26 \ge 26$. True.
$y > \frac{1}{2}x$: $14 > \frac{1}{2}(26) \implies 14 > 13$. True.
$y \le \frac{300-4x}{5}$: $14 \le \frac{300-4(26)}{5} = \frac{196}{5} = 39.2$. True.
So, $(26, 14)$ is a feasible point.

Vertex 2: Intersection of $x=26$ and $y=\frac{300-4x}{5}$.
$x=26 \implies y = \frac{196}{5} = 39.2$.
Since $y$ must be an integer, we take $y=39$. Point $(26, 39)$.
Check $y > \frac{1}{2}x$: $39 > \frac{1}{2}(26) \implies 39 > 13$. True.
So, $(26, 39)$ is a feasible point.

Vertex 3: Intersection of $y=\frac{1}{2}x$ and $y=\frac{300-4x}{5}$.
This intersection is $(\frac{600}{13}, \frac{300}{13}) \approx (46.15, 23.07)$.
We need integer points $(x,y)$ such that $x < 2y$ and $4x+5y \le 300$.
Let's try $x=45$.
Then $y > \frac{1}{2}(45) = 22.5$. So $y$ can be $23, 24, \dots$.
Also $y \le \frac{300-4(45)}{5} = \frac{300-180}{5} = \frac{120}{5} = 24$.
So for $x=45$, $y$ can be 23 or 24.
Consider $(45, 23)$:
$x \ge 26$: $45 \ge 26$. True.
$y > \frac{1}{2}x$: $23 > \frac{1}{2}(45) \implies 23 > 22.5$. True.
$y \le \frac{300-4x}{5}$: $23 \le \frac{300-4(45)}{5} = 24$. True.
So, $(45, 23)$ is a feasible point.

Consider $(45, 24)$:
$x \ge