Maximum and minimum values

Maximum and Minimum Values

Corner Points on the Feasible Region

Example 1

A student has 1200 shillings to spend on exercise books. At the school shop, an exercise book costs 80 shillings, and at a stationery store, it costs 120 shillings. The school shop has only 6 exercise books left, and the student wants to obtain the greatest number of exercise books possible using the money he has. How many exercise books will the student buy from each shop?

Solution

Let:

• $x$ represent the number of books bought from the school shop.
• $y$ represent the number of books bought from the stationery store.

The total money the student has is 1200 shillings, and the cost of a book from the school shop is 80 shillings, while the cost from the stationery store is 120 shillings. The student needs to maximize the total number of books, so we can form the following equations:

• $80x + 120y = 1200$ (the total money constraint),
• $x \leq 6$ (the number of books available at the school shop),
• $x \geq 0, y \geq 0$ (non-negativity constraints).

Using these constraints, we find that the student will buy 6 books from the school shop and the remaining books from the stationery store.

Example 2

A nutritionist prescribes a special diet for patients containing the following number of units of vitamins A and B per kg for two types of food, F1 and F2:

1. F1 provides 10 units of vitamin A and 6 units of vitamin B per kg.
2. F2 provides 12 units of vitamin A and 4 units of vitamin B per kg.

The daily minimum intake required is 120 units of vitamin A and 70 units of vitamin B. What is the least total mass of food a patient must have to meet the daily requirements?

Solution

Let:

• $x$ be the number of kg of food F1 the patient consumes daily.
• $y$ be the number of kg of food F2 the patient consumes daily.

The objective function is to minimize the total mass of food, which is $x + y$.

The constraints are:

• $10x + 12y \geq 120$ (vitamin A constraint),
• $6x + 4y \geq 70$ (vitamin B constraint),
• $x \geq 0, y \geq 0$ (non-negativity constraints).

Solving the system of inequalities, the least total mass of food the patient must have is 6.8 kg.

The Minimum and Maximum Values Using the Objective Function

Example 3

A farmer wants to plant coffee and potatoes. Coffee requires 3 men per hectare, and potatoes also require 3 men per hectare. The farmer has 48 hired laborers available. To maintain a hectare of coffee, the cost is 250 shillings, while the cost for a hectare of potatoes is 100 shillings.

The farmer wants to find the greatest possible land he can sow if he is prepared to use 25,000 shillings.

Solution

Let:

• $x$ be the number of hectares of coffee to be planted.
• $y$ be the number of hectares of potatoes to be planted.

The objective function is to maximize the total hectares, so we want to maximize $x + y$.

The constraints are:

• $3x + 3y \leq 48$ (labor constraint, since 48 workers are available),
• $250x + 100y \leq 25,000$ (cost constraint, since the farmer has 25,000 shillings),
• $x \geq 0, y \geq 0$ (non-negativity constraints).

The inequalities become:

1. $x + y \leq 16$,
2. $5x + 2y \leq 500$.

Now we test the corner points to find the maximum feasible land:

1. At point $A (0, 250)$, the objective function is $f(A) = 250$.
2. At point $B (0, 16)$, the objective function is $f(B) = 16$.
3. At point $C (16, 0)$, the objective function is $f(C) = 16$.
4. At point $D (100, 0)$, the objective function is $f(D) = 100$.

The greatest possible land to be planted is 250 hectares of potatoes.

Note:

In most linear programming problems, non-negativity constraints are included, meaning that $x \geq 0$ and $y \geq 0$. This is because negative quantities do not make sense in practical situations.