C. Solve the following quadratic functions by completing the square method: I. x²+8x−20=0 II. x²+16x=0

Let $C$ be the number of canals for corn.
Let $W$ be the number of canals for wheat.

The objective is to maximize profit $P$.
Profit from corn is Rs. 20 per canal.
Profit from wheat is Rs. 15 per canal.

Constraints:

1. Total land: The farmer possesses 100 canals of land. $$C + W \le 100$$
2. Working hours: Corn requires 3 hours/canal, wheat requires 2 hours/canal. Total working hours cannot exceed 240. $$3C + 2W \le 240$$
3. Non-negativity: The number of canals cannot be negative. $$C \ge 0, W \ge 0$$

The boundary lines are:

1. $C + W = 100$
2. $3C + 2W = 240$
3. $C = 0$ (W-axis)
4. $W = 0$ (C-axis)

The corner points are:

• Intersection of $C=0$ and $W=0$: $(0, 0)$
• Intersection of $C=0$ and $C+W=100$: $(0, 100)$
• Intersection of $W=0$ and $3C+2W=240$: $3C + 2(0) = 240 \Rightarrow 3C = 240 \Rightarrow C = 80$. So, $(80, 0)$.
• Intersection of $C+W=100$ and $3C+2W=240$: From $C+W=100$, we get $W = 100 - C$. Substitute into the second equation: $$3C + 2(100 - C) = 240$$ $$3C + 200 - 2C = 240$$ $$C + 200 = 240$$ $$C = 40$$ Now find $W$: $W = 100 - 40 = 60$. So, $(40, 60)$.

• At $(0, 0)$: $P = 20(0) + 15(0) = 0$
• At $(0, 100)$: $P = 20(0) + 15(100) = 1500$
• At $(80, 0)$: $P = 20(80) + 15(0) = 1600$
• At $(40, 60)$: $P = 20(40) + 15(60) = 800 + 900 = 1700$

The maximum profit is Rs. 1700, which occurs when $C=40$ and $W=60$.

The farmer should cultivate 40 canals of corn and 60 canals of wheat to maximize profit.
The maximum profit is Rs. 1700.