Question

A man has rs.1500 for purchase of rice and wheat. a bag of rice and a bag of wheat cost of rs.180 and rs.120 respectively. he has a storage capacity of 10 bags only. he earns a profit of rs.11 and rs.9 per bag of rice and wheat respectively. formulate an l.p. to maximize the profit and solve it

Answer 1

Answer :

Let, the number of bags of rice and wheat are x and y respectively.

Then, the required LPP be

Max Z = 11x + 9y

Sub to :

180x + 120y ≤ 1500 ...(i)

x + y ≤ 10 ...(ii)

x ≥ 0 and y ≥ 0

We consider (i) and (ii) as equations.

180x + 120y = 1500

➔ x/(25/3) + y/(25/2) = 1 ...(iii)

x + y = 10

➔ x/10 + y/10 = 1 ...(iv)

Now, we consider a set of rectangular axes OX and OY, and plot the lines. Each point has coordinates of the form (x, y). Now, if feasible solution be found, it will be shaded.

(Refer to the attachment) We have found a bounded region OABCO, whose corner points are O (0, 0), A (25/3, 0), B (5, 5) and C (0, 10).

So, Z (at the point O) = 0,

Z (at the point A) = 275/3,

Z (at the point B) = 100 and

Z (at the point C) = 90

So, the required solution is

Max Z = Rs 100 when x = y = 5

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