Question 1148776
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A small furniture shop makes tables and chairs which must be processed through assembly and finishing processes. 
The assembly department is available for 60 hours in every production period while the finishing department is available 
for 48 hours of work.
Making one table requires 4 hours to assemble and 2 hours to finish. 
Each chair requires 2 hours to assemble and 4 hours to finish. 
One table contributes P500 to the profit, while a chair contributes P300. 
Determine the number of tables and chairs to make per production period to maximize profit, assuming all tables and chairs are sold.  
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<pre>
The question is: how many tables (X) and how many chairs (Y) should be produced to maximize the revenue 500*X + 300*Y
under these restrictions:

4X + 2Y <= 60     (1)     (assembly time available) and
2X + 4Y <= 48     (2)     (finishing time available).

In other words, you must maximize the objective function (profit) 

P(X,Y) = 500X + 300Y

over the domain on the plot below, which is  a quadrilateral in the first quadrant (X >= 0,  Y >= 0) restricted 
by the red and the green lines.


    {{{graph( 500, 500, -2, 20, -10, 40,
          (60-4x)/2,  (48-2x)/4
)}}}


    Plots y = {{{(60-4x)/2}}}  (red) and y = {{{(48-2x)/4}}} (green)



The method of linear programming says:

&nbsp;&nbsp;&nbsp;&nbsp;1) Take the vertices of this quadrilateral

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(X1,Y1) = (0,0)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(x2,Y2) = (0,12)  (green line Y-intercept)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(x3,Y3) = (12,6)   (intersection point of the straight lines Y = {{{(60-4x)/2}}} and Y = {{{(48-2x)/4}}} )

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(x4,Y4) = (15,0)   (red line X-intercept)


&nbsp;&nbsp;&nbsp;&nbsp;2) Calculate the objective function at these points

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;R(X1,Y1) = 500*0 + 300*0 = 0.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;R(X2,Y2) = 500*0 + 300*12 = 3600.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;R(X3,Y3) = 500*12 + 300*6 = 7800.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;R(X4,Y4) = 500*15 + 300*0 = 7500.


&nbsp;&nbsp;&nbsp;&nbsp;3) Then select one of these point where the objective function is maximal - In our case this point is (X3,Y3) = (12,6).


&nbsp;&nbsp;&nbsp;&nbsp;4) This point gives your optimal solution X = 12 tables and Y = 6 chairs.


If they follow this optimal solution, their profit will be MAXIMAL, $7800.
</pre>

Solved.


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To see other similar problems solved by the Linear Programming method, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-minimax-problems-by--the-Linear-Programming-method.lesson>Solving minimax problems by the Linear Programming method</A> 

in this site.


Also, &nbsp;look into the solutions of other similar problems in the archive to this forum under the links


https://www.algebra.com/algebra/homework/Inequalities/Inequalities.faq.question.1142000.html 

https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1137172.html

https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1136382.html

https://www.algebra.com/algebra/homework/Coordinate-system/Coordinate-system.faq.question.1134444.html

https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1131906.html

https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1131043.html

https://www.algebra.com/algebra/homework/word/finance/Money_Word_Problems.faq.question.1129285.html

https://www.algebra.com/algebra/homework/playground/test.faq.question.1112482.html

https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1102103.html