Chapter 3.8, Problem 1P

Explanation of Solution

 Formulation of Linear program (LP) to maximize the revenue from candy sales:

 Let “x_ij” be the ounces of ingredient “i” used to make candy “j”.

• Let ingredient 1 is sugar, 2 is nuts, and 3 is chocolate. The candy 1 is slugger candy and 2 is easy out candy.
• The objective of function is to maximize the revenue from the candy sales.
• Each ounce of out candy can be sold for 25 cents and each ounce of slugger candy can be sold for 20 cents.

 Therefore, the objection function is,

 $\text{Maximize, }z=25\left(x_{12}+x_{22}+x_{32}\right)+20\left(x_{11}+x_{21}+x_{31}\right)$

 Constraint 1:

 At present, 100 oz of sugar is in the stock.

 $\begin{array}{c}\text{Ounces of sugar in slugger candy + ounces of sugar in easy out candy }\le \text{ 100}\\ x_{11}+x_{12}\le 100\end{array}$

 Constraint 2:

 At present, 20 oz of nuts is in the stock.

 $\begin{array}{c}\text{Ounces of nuts in slugger candy + ounces of nuts in easy out candy }\le \text{ 20}\\ x_{21}+x_{22}\le 20\end{array}$

 Constraint 3:

 At present, 30 oz of nuts is in the stock.

 $\begin{array}{c}\text{Ounces of chocolate in slugger candy + ounces of chocolate in easy out candy }\le \text{ 30}\\ x_{31}+x_{32}\le 30\end{array}$

 Constraint 4:

 At least, 20% of nuts must be there in easy out candy.

 $20\%\text{ of nuts in easy out candy }\ge \text{0}\text{.2}$

 $x_{22}\ge \text{0}\text{.2}\left(x_{12}+x_{22}+x_{32}\right)$

 Constraint 5:

 At least, 10% of nuts must be there in slugger candy.

 $10\%\text{ of nuts in slugger candy }\ge \text{0}\text{.1}$

 $x_{21}\ge \text{0}\text{.1}\left(x_{11}+x_{21}+x_{31}\right)$

 Constraint 6:

 At least, 10% of chocolate must be there in slugger candy