Question

A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

kg per bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Answer 1

Let x kg and y kg of brand P and brand Q respectively of fertilizer be used.
Quantity of fertilizer cannot be negative.
Therefore,

kg per bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

It is given that the fruit grower tends to minimise the amount of nitrogen.

​Therefore, Z = 3x + 3.5y

Garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

The constraints are :

$$\begin{gathered} x+2y\ge 240 \\ 3x+1.5y\ge 270 \\ 1.5x+2y\ge 310 \end{gathered}$$

The mathematical form of the given LPP is:

Min Z = 3x + 3.5y

subject to constraints:

$$\begin{gathered} x+2y\ge 240 \\ 3x+1.5y\ge 270 \\ 1.5x+2y\ge 310 \end{gathered}$$

First we will convert inequations into equations as follows:
x + 2y = 240, 3x + 1.5y = 270, 1.5x + 2y =310, x = 0 and y = 0

Region represented by x + 2y ≥ 240:
The line x + 2y = 240 meets the coordinate axes at A(240, 0) and B(0, 120) respectively. By joining these points we obtain the line x + 2y = 240 . Clearly (0,0) does not satisfies the x + 2y = 240. So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 240.

Region represented by 3x + 1.5y ≥ 270:
The line 3x + 1.5y = 270 meets the coordinate axes at C(90, 0) and D(0, 180) respectively. By joining these points we obtain the line 3x + 1.5y = 270. Clearly (0,0) does not satisfies the inequation 3x + 1.5y ≥ 270. So,the region which does not contain the origin represents the solution set of the inequation
3x + 1.5y ≥ 270.

Region represented by 1.5x + 2y ≥ 310:
The line 1.5x + 2y = 310 meets the coordinate axes at $E\left(\frac{3100}{15},0\right)$ and F(0, 155) respectively. By joining these points we obtain the line 1.5x + 2y = 310 . Clearly (0,0) does not satisfies the inequation 1.5x + 2y ≥ 310. So,the region which does not contain the origin represents the solution set of the inequation 1.5x + 2y ≥ 310.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≥ 310, x ≥ 0 and y ≥ 0 are as follows:

In the above graph, the shaded region is the feasible region. The corner points are D(0, 180), G(20, 140), H(140, 50) and A(240, 0).

The values of the objective function Z at corner points of the feasible region are given in the following table:

Corner Points	Z = 3x + 3.5y
D(0, 180)	630
G(20, 140)	550
H(140, 50)	595
A(240, 0)	720

Clearly, Z is minimium x = 20 and y = 140 and the minimum value of Z at this point is 550.

Thus, 20 kg of brand P and 140 kg of brand Q fertilizer should be used to minimise the amount of nitrogen which is 550 kg.