Chapter 3.8, Problem 6P

Explanation of Solution

 Formulation of a Linear Programming (LP) to help Bullco maximize the profit:

 Let consider “x_s1” and “x_s2” be the number of pounds of silicon used in fertilizer 1 and fertilizer 2.

 Let “x_n1” and “x_n2” be the number of pounds of nitrogen used in fertilizer 1 and fertilizer 2.

 The objective is to maximize the profits of Bullco.

 $\begin{array}{c}\text{z}=\left(\begin{array}{l}\left(\text{selling price of fertilizer 1}\right)\left(\text{amount of silicon and nitrogen used to produce fertilizer 1}\right)+\\ \left(\text{selling price of fertilizer 2}\right)\left(\text{amount of silicon and nitrogen used to produce fertilizer 2}\right)-\\ \left(\text{cost of silicon}\right)\left(\text{amount of silicon used to produce fertilizer 1 and 2}\right)-\\ \left(\text{cost of nitrogen}\right)\left(\text{amount of nitrogen used to produce fertilizer 1 and 2}\right)\end{array}\right)\\ =70\left(x_{s1}+x_{n1}\right)+40\left(x_{s2}+x_{n2}\right)-10\left(x_{s1}+x_{s2}\right)-15\left(x_{n1}+x_{n2}\right)\end{array}$

 Therefore, the objective function is,

 Maximum $z=70\left(x_{s1}+x_{n1}\right)+40\left(x_{s2}+x_{n2}\right)-10\left(x_{s1}+x_{s2}\right)-15\left(x_{n1}+x_{n2}\right)$

 Constraint 1:

 At most, 100lb of silicon can be purchased.

 Amount of silicon used to produce fertilizer 1 and 2 $\le 100$

 $x_{n1}+x_{n2}\le 100$

 Constraint 2:

 At most, 80lb of nitrogen can be purchased