In: Finance
A company blends nitrogen and phosphorous to produce two types of fertilizers. Fertilizer 1 must be at least 50% nitrogen and sells for $55 per pound. Fertilizer 2 must be at least 55% phosphorous and sells for $45 per pound. The company can purchase up to 9000 pounds of nitrogen at $20 per pound and up to 12,000 pounds of phosphorous at $12 per pound.
Assuming that all fertilizer produced can be sold, determine the optimal blending plan for the company. What is the maximum profit?
Nitrogen | Phosphorous | Price | Cost | Profit | |
Fertilizer 1 | 50% | 50% | 55 | 16 | 39 |
Fertilizer 2 | 45% | 55% | 45 | 15.6 | 29.4 |
Price (Per Pound) | 20 | 12 | |||
Max Quantity | 9000 | 12000 | |||
Let Quantity of Fert 1 | x | ||||
Let Quantity of Fert 2 | y | ||||
Therefore, | |||||
Maximize | 39x+29.4y | -eq (3) | |||
s.t. | |||||
0.5x+0.45y | <= | 9000 | -eq (1) | ||
0.5x+ 0.55 y | <= | 12000 | -eq (2) | ||
From Above Equation: | Nitrogen is the limiting variable | ||||
This means that eq(2) is redundant | |||||
Now maximizing eq(3) w.r.t. eq (1) | |||||
Slope of eq (1) | -1.11111 | ||||
Slope of eq (3) | -1.32653 | ||||
Since, Eq (3) is steeper than Eq (1), then we can say that profit will be maximized at y=0, i.e. by producing fertilizer 1 only. | |||||
Therefore, | |||||
x | = | 18000 | |||
y | = | 0 | |||
Profit | = | 702000 |