Graphical solution

Let $x$ be number of exercise books to be bought at school shop

Let $y$ be number of exercise books to be bought at stationery shop

Objective function

Let $f(x, y)$ = objective function

Then $f(x, y) = x + y$

Constraints or linear inequalities

$8x + 12y \leq 120$

$x \leq 6$

Non-negative constraints:

$x \geq 0$

$y \geq 0$

Equations

$8x + 12y = 120$

$4x + 6y = 60$

$2x + 3y = 30$

When $x = 0$, $y = 10$

$y = 0$, $x = 15$

$x = 6$, $y = 0$

$x = 6$

The greatest numbers of exercise books he can buy are 12 books: 6 from the school shop and 6 from stationery.

Students in a certain class are about to take a certain test of BAM which has two sections A and B; where in section A each question worth 10 marks while in section B each worth 25 marks. The student must do at least 3 questions of section A but not more than 12. A student must also do 4 questions from section B but not more than 15. In addition students cannot do more than 20 questions. How many questions of each type should the student do to obtain the maximum scores?

Let $x$ be number of questions to be done in section A

Let $y$ be number of questions to be done in section B

Objective function

$f(x, y) = 10x + 25y$

Constraints

$3 \leq x \leq 12$

$4 \leq y \leq 15$

$x + y \leq 20$

$x \geq 0$, $y \geq 0$

Maximize $f(x, y) = 10x + 25y$ subject to:

$3 \leq x \leq 12$

$4 \leq y \leq 15$

$x + y \leq 20$

$x \geq 0$, $y \geq 0$

Equations

$x = 3$ or $x = 12$

$y = 4$ or $y = 15$

$x + y = 20$

$x = 20$, $y = 0$

$x = 0$, $y = 20$

$x = 0$

$y = 0$

The student should do 5 questions from section A and 15 questions from section B to obtain maximum score of 425.

Example 01

A doctor prescribes a special diet for patients containing the following number of units of Vitamin A and B per kg of two types of food F1 and F2.

If the minimum daily intake required is 120 units of A and 70 units of B, what is the least total mass of food a patient must have so as to have enough of these vitamins?

Example 02

Rice and beans provide maximum levels of protein, calories and vitamin B2. If used as a staple diet. The food values per kg of uncooked rice and beans are as shown in the table below.

What is the lowest cost of diet meeting these specifications?

Let $x$ be number of kg of rice to be bought

Let $y$ be number of kg of beans to be bought

Objective function

Minimize $f(x, y) = 400x + 500y$

Constraints

$60x + 90y \geq 120$

$3200x + 1000y \geq 2000$

$0.4x + 0.1y \geq 0.2$

$x \geq 0$, $y \geq 0$

For $60x + 90y \geq 120$:

$60x + 90y = 120$

$2x + 3y = 4$

When $x = 0$, $y = 1.33$

$y = 0$, $x = 2$

For $3200x + 1000y \geq 2000$:

$32x + 10y = 20$

$16x + 5y = 10$

When $x = 0$, $y = 2$

$y = 0$, $x = 0.625$

For $0.4x + 0.1y \geq 0.2$:

$0.4x + 0.1y = 0.2$

When $x = 0$, $y = 2$

When $y = 0$, $x = 0.5$

The least total mass a patient should have is 6.8 kg, i.e. 3.6 kg of food 1 and 3.2 kg of food 2.