Answer to Question #280703 in Operations Research for Jel

Question

A factory produces three types of fertilizer: X1, X2 and X3. The profits realized from a kilo of each typeare P25, P15 and P35, respectively. The raw materials from which fertilizers are made are nitrogen, sulfurand potassium which are used in the following quantities:NitrogenSulfurPotassiumX13 kilos4 kilos8 kilosX23 kilos4 kilos1 kiloX32 kilos5 kilos3 kilosThe available stocks include 1,500 kilos of nitrogen, 1000 kilos of sulfur and 1200 kilos of potassium. Theobjective is to maximize profit. Find the amounts of X1, X2 and X3 to be manufactured

Answer 1

Solution:

Given the information, we can write the equation for profit as: 50X + 75Y

We are given the constraints as:

For nitrogen, 4X + 3Y $\leq$ 1500 or X/375 + Y/500 $\leq$ 1

For sulfur, 4X + 2Y $\leq$ 1200 or X/300 + Y/600 $\leq$ 1

For potassium, 9X + Y $\leq$ 1200 or X/133.33 + Y/1200 $\leq$ 1

The required points of common area would be: origin (0, 0), x-intercept of potassium constraint (1200/9, 0), y-intercept of nitrogen constraint (0, 500), and intersection point between nitrogen and potassium constraints:

4X + 3Y = 1500

X = (1500 - 3Y)/4

Substituting in potassium constraint, we get: 9(1500 - 3Y)/4 + Y = 1200

3375 - 31Y/4 = 1200

Y = (3375 - 1200)(4/31) = 8700/31

X = (1500 - 3*8700/31)/4 = 5100/31 (sulfur constraint is satisfied automatically with above case)

Then, we can find the combination (of the above 4) generating maximum profit as:

At (0, 0), profit = 50*0 + 75*0 = $0

At (1200/9, 0), profit = 50*1200/9 + 75*0 = $6,666.67

At (0, 500), profit = 50*0 + 75*500 = $37,500

At (5100/31, 8700/31), profit = 50*5100/31 + 75*8700/31 = (255000 + 652500)/31 = $29,274.19

Thus, the amounts of (X, Y) = (0, 500) should be produced in order to maximize the profit.