Chapter 3, Problem 36RP

a.

Program Plan Intro

Linear program:

Let,

$\text{x}$ be the number of layout in Type 1

$\text{y}$ be the number of layout in Type 2

$\text{z}$ be the number of layout in Type 3

The, the LPP becomes,

Maximize, $\text{Z=50x+30y+60z}$

Subject to the constraints,

$\begin{array}{l}\text{Z=50x+30y+60z}\\ \text{30x+10y+20z}\le \text{1000}\\ \text{20x+40y+50z}\le \text{800}\\ \text{4x+3y+32z}\le \text{100}\end{array}$

And $\text{x,y,z}\ge \text{0}$

This is the required linear programming problem.

Explanation of Solution

b.

Change to the LP:

Given, there must be at least 5 layouts in each type.

We have,

$\text{x}$ be the number of layout in Type 1

$\text{y}$ be the number of layout in Type 2

$\text{z}$ be the number of layout in Type 3

Then the constraint,

$\text{x}\ge 5\text{,y}\ge 5\text{,z}\ge 5$ should be added to the LP.

Then the LPP becomes,

Maximize, $\text{Z=50x+30y+60z}$

Subject to the constraints,

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