Question

A fertilizer company blends silicon and nitrogen to produce two types of fertilizers. Fertilizer 1 must be at least 40% nitrogen and sells for $70 per pound. Fertilizer 2 must be at least 70% silicon and sells for $40 per pound. The company can purchase up to 8000 pounds of nitrogen at $15 per pound and up to 10,000 pounds of silicon at $10 per pound. a. Assuming that all fertilizer produced can be sold, determine how the company can maximize its profit.b. Use SolverTable to explore the effect on profit of changing the minimum percentage of nitrogen required in fertilizer 1. c. Suppose the availabilities of nitrogen and silicon both increase by the same percentage from their current values. Use SolverTable to explore the effect of this change on profit.

Answer 1

Maximize

z = 70(X$s_{1}$ + X$n_{1}$ ) + 40(X$s_{2}$ + X$n_{2}$ ) - 10 (X$s_{1}$ + X$s_{2}$ ) - 15(X$n_{1}$ + X$n_{2}$ )  

Subject to the constraints  

X$s_{1}$ + X$s_{2}$ ≤ 10000

X$n_{1}$ + X$n_{2}$ ≤ 8000

X$n_{1}$ ≥ 0.4 ( X$s_{1}$ + X$n_{1}$)

X$s_{2}$ ≥ 0.7 ( X$s_{2}$ + X$n_{2}$ )

All Variables ≥ 0

Firstly let us consider X$s_{1}$ and X$s_{2}$ to be the number of pounds of silicon used in fertilizer1 and fertilizer2 respectively

Also, let X$n_{1}$ and X$n_{2}$ be the number of pounds of nitrogen used in fertilizer1 and fertilizer2 respectively

We know that the objective is to maximize profits.

z = [(Selling price of fertilizer1) (Amount of silicon and nitrogen used to produce fertilizer1) + (Selling price of fertilizer2) (Amount of silicon and nitrogen used to produce fertilizer2) - (Cost of silicon) (Amount of silicon used to produce fertilizer I and 2) - (Cost of nitrogen) (Amount of nitrogen used to produce fertilizer I and 2)]

z= 70 (X$s_{1}$ + X$n_{1}$ ) + 40 (X$s_{2}$ + X$n_{2}$) - 10 (X$s_{1}$ + X$s_{2}$) - 15( X$n_{1}$ + X$n_{2}$ )

Now

Constraint 1;  At most, 100 lb of silicon can be purchased

Amount of silicon used to produce fertilizer 1 and 2 ≤ 10000

X$s_{1}$ + X$s_{2}$ ≤ 10000

Constraint 2; At most, 80 lb of nitrogen can be purchased

Amount of nitrogen used to produce fertilizer 1 and 2 ≤ 8000

X$n_{1}$ + X$n_{2}$ ≤ 8000

Constraint 3; Fertilizer 1 must be at least 40% of nitrogen

Amount of nitrogen used to produce fertilizer 1 ≥ 40% (fertilizer 1)

X$n_{1}$ ≥ 0.4 ( X$s_{1}$ + X$n_{1}$ )

Constraint 4; Fertilizer 2 must be at least 70% of silicon

Amount of silicon used to produce fertilizer 2  ≥ 70% (fertilizer 2)

X$s_{2}$ ≥ 0.7 ( X$s_{2}$ + X$n_{2}$ )

so the formalization of the given linear program is,  

Maximize

z = 70(X$s_{1}$ + X$n_{1}$ ) + 40(X$s_{2}$ + X$n_{2}$ ) - 10 (X$s_{1}$ + X$s_{2}$ ) - 15(X$n_{1}$ + X$n_{2}$ )  

Subject to the constraints  

X$s_{1}$ + X$s_{2}$ ≤ 10000

X$n_{1}$ + X$n_{2}$ ≤ 8000

X$n_{1}$ ≥ 0.4 ( X$s_{1}$ + X$n_{1}$)

X$s_{2}$ ≥ 0.7 ( X$s_{2}$ + X$n_{2}$ )

All Variables ≥ 0

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