Chapter 6.9, Problem 2P

a.

Explanation of Solution

Profit value that makes the current basis remains optimal:

• Let “${\text{c}}_1$” be the profit of the candy bar of Type “1”
• The constraint of the dual for “${\text{x}}_1$” is as follows,

${\text{y}}_{\text{1}}{\text{+ 2y}}_{\text{2}}\ge {\text{c}}_{\text{1}}$

• The shadow prices for the problem is,

$\begin{array}{l}{\text{y}}_{\text{1}}\text{=4}\\ {\text{y}}_{\text{2}}\text{=1}\end{array}$

b.

Explanation of Solution

“No”, the current basis is no longer optimal, if the candy bar of Type “1” used “0.5” oz sugar, and “0.75” oz of chocolate.

Reasons:

• The constraint of the dual for the candy bar of Type “1” is as follows,

$\text{0}{\text{.5y}}_{\text{1}}+0.75{\text{y}}_2\ge 3$

• The shadow prices for the problem is,

$\begin{array}{l}{\text{y}}_{\text{1}}\text{=4}\\ {\text{y}}_{\text{2}}\text{=1}\end{array}$

c.

Explanation of Solution

“No”, the current basis is no longer optimal, if the candy bar of Type “4” produces a profit of “$\text{10}\cancel{\text{c}}$” and utilizes “2” oz of sugar and “1” oz of chocolate.

Reasons:

• The constraint of the dual for the candy bar of Type “4” is as follows,

${\text{2y}}_{\text{1}}+{\text{y}}_2\ge 10$

• The shadow prices for the problem is,

$\begin{array}{l}{\text{y}}_{\text{1}}\text{=4}\end{array}$