A cow needs 160 units of Vitamin A and 40 units of Vitamin B. Feed type I contains 20 units of Vitamin A and 4 units of Vitamin B per kg. Feed type II contains 10 units of Vitamin A and 10 units of Vitamin B per kg. If x kg of feed type I and y kg of feed type II are used, form the inequalities and determine the least cost of feed to meet the minimum requirements.

Let $x$ be the number of kg of Type X feed.
Let $y$ be the number of kg of Type Y feed.

The constraints are based on the minimum vitamin requirements:

• Vitamin A requirement: The cow needs at least 160 units of Vitamin A. Type X provides 9 units/kg, and Type Y provides 20 units/kg. $$9x + 20y \ge 160$$
• Vitamin B requirement: The cow needs not less than 160 units of Vitamin B. Type X provides 4 units/kg, and Type Y provides 40 units/kg. $$4x + 40y \ge 160$$ This inequality can be simplified by dividing by 4: $$x + 10y \ge 40$$
• Non-negativity constraints: The amount of feed cannot be negative. $$x \ge 0$$ $$y \ge 0$$

The objective function to minimize is the total cost $C$:

• Cost: Type X feed costs 80 units/kg, and Type Y feed costs 100 units/kg. $$C = 80x + 100y$$

The inequalities are:

• $9x + 20y \ge 160$
• $x + 10y \ge 40$
• $x \ge 0$
• $y \ge 0$

To graph the inequalities, we first consider the boundary lines:

1. $L_1: 9x + 20y = 160$

   • If $x = 0$, $20y = 160 \implies y = 8$. Point: $(0, 8)$
   • If $y = 0$, $9x = 160 \implies x = \frac{160}{9} \approx 17.78$. Point: $(\frac{160}{9}, 0)$ The region $9x + 20y \ge 160$ is above or to the right of this line.

2. $L_2: x + 10y = 40$

   • If $x = 0$, $10y = 40 \implies y = 4$. Point: $(0, 4)$
   • If $y = 0$, $x = 40$. Point: $(40, 0)$ The region $x + 10y \ge 40$ is above or to the right of this line.

The feasible region is the area in the first quadrant ($x \ge 0, y \ge 0$) that satisfies both $9x + 20y \ge 160$ and $x + 10y \ge 40$. This region is unbounded.

To find the corner points of the feasible region, we find the intersection of the boundary lines:

• Intersection of $L_1$ and $L_2$: From $x + 10y = 40$, we have $x = 40 - 10y$. Substitute into $9x + 20y = 160$: $9(40 - 10y) + 20y = 160$ $360 - 90y + 20y = 160$ $360 - 70y = 160$ $70y = 200$ $y = \frac{200}{70} = \frac{20}{7}$ Now find $x$: $x = 40 - 10\left(\frac{20}{7}\right) = 40 - \frac{200}{7} = \frac{280 - 200}{7} = \frac{80}{7}$ Intersection point: $\left(\frac{80}{7}, \frac{20}{7}\right)$

The corner points of the feasible region are:

• $(0, 8)$ (intersection of $L_1$ and the y-axis)
• $\left(\frac{80}{7}, \frac{20}{7}\right)$ (intersection of $L_1$ and $L_2$)
• $(40, 0)$ (intersection of $L_2$ and the x-axis)

Graph description:
Draw an x-axis and a y-axis.
Plot the line $9x + 20y = 160$ by connecting $(0, 8)$ and $(\frac{160}{9}, 0) \approx (17.78, 0)$.
Plot the line $x + 10y = 40$ by connecting $(0, 4)$ and $(40, 0)$.
The feasible region is the area in the first quadrant ($x \ge 0, y \ge 0$) that lies above both lines. This region is bounded by the y-axis from $y=8$ upwards, by the line $9x+20y=160$ from $(0,8)$ to $(\frac{80}{7}, \frac{20}{7})$, by the line $x+10y=40$ from $(\frac{80}{7}, \frac{20}{7})$ to $(40,0)$, and by the x-axis from $x=40$ rightwards.

To find the least cost, we evaluate the objective function $C = 80x + 100y$ at each corner point of the feasible region:

• At $(0, 8)$:

  $$C = 80(0) + 100(8) = 0 + 800 = 800$$

• At $\left(\frac{80}{7}, \frac{20}{7}\right)$:

  $$C = 80\left(\frac{80}{7}\right) + 100\left(\frac{20}{7}\right) = \frac{6400}{7} + \frac{2000}{7} = \frac{8400}{7} = 1200$$

• At $(40, 0)$:

  $$C = 80(40) + 100(0) = 3200 + 0 = 3200$$

Comparing the costs, the minimum cost is 800.

The least cost of maintaining a cow at Jimmy's farm is 800 units, achieved by using 0 kg of Type X feed and 8 kg of Type Y feed.

a) The inequalities are:

$$9x + 20y \ge 160$$ $$x + 10y \ge 40$$ $$x \ge 0$$ $$y \ge 0$$

b) The feasible region is the area in the first quadrant bounded by the lines $9x + 20y = 160$ and $x + 10y = 40$, lying above both lines. The corner points of this region are