UPSC 2018 Maths Optional Paper 2 Q1e — Step-by-Step Solution

10 marks · Section A

LPP: standard form; basic, basic feasible, optimal solutions · Linear Programming · asked 6× in 13 yrs · Read the full method →

Question

An agricultural firm has 180 tons of nitrogen fertilizer, 250 tons of phosphate and 220 tons of potash. It will be able to sell a mixture of these substances in their respective ratio $3:3:4$ at a profit of Rs. 1500 per ton and a mixture in the ratio $2:4:2$ at a profit of Rs. 1200 per ton. Pose a linear programming problem to show how many tons of these two mixtures should be prepared to obtain the maximum profit.

Technique

Translate ratios into per-ton resource coefficients (divide each ratio by the sum of its parts), then write a max-profit LPP with $\le$ availability constraints.

Solution

Step 1 — Decision variables

Let

x_1=\text{tons of Mixture I prepared (ratio }3:3:4),\qquad x_2=\text{tons of Mixture II prepared (ratio }2:4:2).

Step 2 — Resource content per ton of each mixture

A ratio $3:3:4$ has parts summing to $3+3+4=10$ . So per ton of Mixture I:

\text{nitrogen }=\tfrac{3}{10},\quad \text{phosphate }=\tfrac{3}{10},\quad \text{potash }=\tfrac{4}{10}.

A ratio $2:4:2$ has parts summing to $2+4+2=8$ . So per ton of Mixture II:

\text{nitrogen }=\tfrac{2}{8}=\tfrac14,\quad \text{phosphate }=\tfrac{4}{8}=\tfrac12,\quad \text{potash }=\tfrac{2}{8}=\tfrac14.

Step 3 — Objective function

Profit is Rs. $1500$ per ton of Mixture I and Rs. $1200$ per ton of Mixture II:

\text{Maximize } Z=1500\,x_1+1200\,x_2.

Step 4 — Resource (availability) constraints

Total of each substance used cannot exceed the stock (180 N, 250 phosphate, 220 potash):

\begin{aligned} \text{Nitrogen: }&\ \tfrac{3}{10}x_1+\tfrac14 x_2\le 180,\\ \text{Phosphate: }&\ \tfrac{3}{10}x_1+\tfrac12 x_2\le 250,\\ \text{Potash: }&\ \tfrac{4}{10}x_1+\tfrac14 x_2\le 220,\\ &\ x_1,x_2\ge0. \end{aligned}

Step 5 — Cleared-fraction (integer-coefficient) form

Multiplying each constraint by $20$ to clear denominators gives an equivalent, tidier statement:

Answer

\boxed{\; \begin{aligned} \text{Maximize }&\ Z=1500\,x_1+1200\,x_2\\[2pt] \text{s.t. }&\ 6x_1+5x_2\le 3600\quad(\text{nitrogen}\times20)\\ &\ 6x_1+10x_2\le 5000\quad(\text{phosphate}\times20)\\ &\ 8x_1+5x_2\le 4400\quad(\text{potash}\times20)\\ &\ x_1,x_2\ge 0. \end{aligned}\;}