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December 16th, 2014, 02:31 PM   #1
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linear programming problem 2

There are 6 pieces of wooden boards of a furniture craftsman and has 28 hours of free time. Sells two types of libraries by the board. Type I library requires 2 wooden boards and 7 hours; The library requires a type II 1 plate and 8 hours. The selling price of 120,000 \$ Type I library. The price is 80,000 \$ type II library. According to want to maximize sales revenue Furnishers must be purchased from both models make how many of each?
Install problem as a linear programming model.

Last edited by greg1313; December 16th, 2014 at 02:34 PM.
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December 17th, 2014, 07:13 PM   #2
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What does "plate" mean?
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December 17th, 2014, 10:18 PM   #3
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Hello, kalli1!

Your English is not too accurate.
I'll modify the wording for simplicity.


Quote:
A furniture craftsman has 6 boards and 28 hours of time.
Type A tables require 2 boards and 7 hours of labor.
Type B tables require 1 board and 8 hours of labor.
Type A tables sell for \$1200; type B tables sell for \$800.

How many of each type of table will maximize sales revenue?

Let $x$ = number of type A tables: $\:x \ge 0$
Let $y$ = number of type B tables: $\:y \ge 0$

We have:

$\quad \begin{array}{c|c|c|c|} & \text{boards} & \text{hours} & \text{Revenue} \\ \hline
A & 2x & 7x & 1200x \\
B & y & 8y & 800y \\ \hline
& 6 & 28 \\ \hline \end{array}$


In Quadrant I, we must graph and shade:

$\quad\begin{array}{ccc} 2x + y & \le & 6 \\ 7x + 8y &\le & 28 \end{array}$

Then find the coordinates of the vertices of the shaded area.

Then substitute them into the Revenue function: $\:R \;=\;1200x+800y$
$\quad$ and determine which produces maximum revenue.

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December 18th, 2014, 04:36 AM   #4
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sorry for my english and thank you so much.
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