In the previous section, we saw the mathematical model for the motion of a simple pendulum. In this section, we will see another example.
Example 3:
•
The situation is:
A farm house uses atleast 800 kg of special food daily. The special food is a mixture of corn and soyabean with the following compositions:
$\begin{array}{cc}{} &{\textbf{Material}} &{\textbf{Protein present per Kg}} &{\textbf{Fibre present per Kg}} &{\textbf{Cost per Kg}}\\
{} &{\textbf{Corn}} &{0.09} &{0.02} &{\text{Rs 10}}\\
{} &{\textbf{Soyabean}} &{0.60} &{0.06} &{\text{Rs 20}}\\
\end{array}$
The dietary requirements of the special food stipulate atleast 30% protein and at most 5% fibre. How can this dietary requirement be satisfied at the least possible cost?
•
This situation can be converted into a mathematical problem. For that,
we engage in the process of mathematical modeling. The final result can
be obtained in 4 steps:
Step 1:
First we
study the situation. We will write it in order:
(i) “Atleast 800 kg” means:
♦ The total quantity should not be less than 800 kg.
♦ The total quantity can be greater than 800 kg.
(ii) The available nutrients and costs can be listed:
• One kg corn will
♦ give 0.09 kg protein.
♦ give 0.02 kg fibre.
♦ cost Rs 10
• One kg soya bean will give
♦ give 0.60 kg protein.
♦ give 0.06 kg fibre.
♦ cost Rs 20
(iii) Our aim is:
• The 800 kg (or more) food must
♦ contain atleast 30% protein.
♦ contain at most 5% fibre.
♦ cost the least possible amount.
•
We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.
Step 2:
This can be written in order:
(i) Let the food be made up of x kg corn and y kg of soya bean.
• Then from (i) of step 1, we can write: x + y ≥ 800
(ii) Available protein and fibre:
• From the x kg corn, we will get some protein and fibre.
• From the y kg soya bean also, we will get some protein and fibre.
• Then from (ii) of step 1, we can write:
♦ Total protein than can be obtained from x kg corn and y kg soya bean is:
0.09x + 0.60y
♦ Total fibre than can be obtained from x kg corn and y kg soya bean is:
0.02x + 0.06y
(iii) Checking whether the dietary requirements are fulfilled:
• We are taking (x+y) kg food. This will contain (0.09x + 0.60y) kg of protein.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{} &{\frac{0.09x + 0.60y}{x+y} \times 100} & {~\ge~} &{30} &{} \\
{\Rightarrow} &{\frac{0.09x + 0.60y}{x+y}} & {~\ge~} &{0.30} &{} \\
{\Rightarrow} &{0.09x + 0.60y} & {~\ge~} &{0.30 x + 0.30y} &{} \\
{\Rightarrow} &{0.30y} & {~\ge~} &{0.21 x} &{} \\
{\Rightarrow} &{0} & {~\ge~} &{0.21 x – 0.30 y} &{} \\
{\Rightarrow} &{0.21 x – 0.30 y} & {~\le~} &{0} &{} \\
\end{array}$
• We are taking (x+y) kg food. This will contain (0.02x + 0.06y) kg of fibre.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{} &{\frac{0.02x + 0.06y}{x+y} \times 100} & {~\le~} &{5} &{} \\
{\Rightarrow} &{\frac{0.02x + 0.06y}{x+y}} & {~\le~} &{0.05} &{} \\
{\Rightarrow} &{0.02x + 0.06y} & {~\le~} &{0.05 x + 0.05y} &{} \\
{\Rightarrow} &{0.01y} & {~\le~} &{0.03 x} &{} \\
{\Rightarrow} &{0} & {~\le~} &{0.03 x – 0.01 y} &{} \\
{\Rightarrow} &{0.03 x – 0.01 y} & {~\ge~} &{0} &{} \\
\end{array}$
(iv) Checking cost:
We are taking (x+y) kg food. That food will cost (10x + 20y)
Then from (iii) of step 1, we can write: (10x + 20y) should be as small as possible.
•
We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations/inequalities. In short, we write a mathematical problem
in this step. Any mathematical problem will have a definite solution.
So in this step, it is necessary to recheck all the works done thus far.
Step 3:
In this step, we put the equations and inequalities into actual use. This can be written in order:
(i) From (i) of step 2, we have:
x + y ≥ 800
• The red line in fig.B.2 below represents
x + y = 800
 |
| Fig.B.2 |
• So any (x,y) in the upper half plane II of the red line will satisfy the inequality. See section 6.3.
(ii) From (iii) of step 2, we have:
0.21x - 0.30y ≤ 0
• The magenta line in fig.B.2 above represents
0.21x - 0.30y = 0
• So any (x,y) in the upper half plane II of the magenta line will satisfy the inequality.
(iii) Again from (iii) of step 2, we have:
0.03x - 0.01y ≥ 0
• The green line in fig.B.2 above represents
0.03x - 0.01y = 0
• So any (x,y) in the lower half plane I of the green line will satisfy the inequality.
(iv) So we have three regions:
♦ upper half plane II of the red line.
♦ upper half plane II of the magenta line.
♦ lower half plane I of the green line.
• The three regions will over lap in the yellow region shown in fig.B.2.
• So any (x,y) in the yellow region will satisfy all the three inequalities.
(v) From (iv) of step 2, we have:
(10x + 20y) must be as small as possible.
• For that, x and y must be as small as possible. At the same time, x and y must satisfy all three inequalities.
• This point is marked by a cyan dot in fig.B.2 above.
• If we draw a vertical line through the cyan dot, then that line will meet the x-axis at 470.59
• If we draw a horizontal line through the cyan dot, then that line will meet the y-axis at 329.41
• So the required point is (470.6,329.4)
•
We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for
lengthy problems.
Step 4:
This can be written in order:
(i) First we do the validation.
• We are going to buy 470.6 kg corn and 329.4 kg soya bean.
• So total quantity = (470.6 + 329.4) = 800.
Thus “atleast 800” is satisfied.
• In this way all the dietary requirements can be checked. This will complete the validation process.
(ii) Next we do the interpretation.
• We state that:
If we buy 470.6 kg corn and 329.4 kg soya bean, then all the dietary requirements will be satisfied and at the same time, the cost will be the least possible value. The least possible cost is (470.6 × 10 + 329.4 × 20) = Rs 11294.00
• We can write:
In step 4, we do validation and interpretation.
In the next section, we will see one more example.
