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The Natural Fertilizer Company needs to determine the number of 100-pound sacks of two types of fertilizer to produce in order to maximize profit.

- **Fertilizers:**
- **Lawn Fertilizer (35-20-20):** 35% nitrate, 20% phosphate, 20% potash
- **Garden Fertilizer (14-13-12):** 14% nitrate, 13% phosphate, 12% potash

- **Resources Available:**
- 8 tons of nitrate
- 11 tons of phosphate
- 7 tons of potash

- **Profit:**
- $15.00 per sack of lawn fertilizer
- $10.00 per sack of garden fertilizer

**Objective:**
Maximize profit.

**Variables:**
- Let [tex]x[/tex] be the number of 100-pound sacks of lawn fertilizer.
- Let [tex]y[/tex] be the number of 100-pound sacks of garden fertilizer.

**Constraints:**
1. Nitrate: \([tex]0.35x + 0.14y \leq 8000[/tex]\) (in pounds)
2. Phosphate: \([tex]0.20x + 0.13y \leq 11000[/tex]\) (in pounds)
3. Potash: \([tex]0.20x + 0.12y \leq 7000[/tex]\) (in pounds)
4. Non-negativity: \([tex]x \geq 0[/tex]\), \([tex]y \geq 0[/tex]\)

**Objective Function:**
Maximize profit: \([tex]15x + 10y[/tex]\)

Formulate this linear programming problem using the information provided above.

Answer :

Answer:

Step-by-step explanation:

Here the number of variables is 2

Variable x should be the number of 100 pound bags of lawn fertilizer to be made.

Variable y should be the number of 100 pound bags of garden fertilizer to be made.

The number of constraints is 5

They are,

0.35x + 0.14y ≤ 160 ----(1)

0.20x + 0.13y ≤ 220 ----(2)

0.20x + 0.12y ≤ 140 ----(3)

x ≥ 0 ----(4)

y ≥ 0 ----(5)

The number of objective functions is 1

It is, Maximize Z = 15x + 10y

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