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This a link to the question, it includes a graph as well. 

 

https://free-picload.com/image/qtfKRh

 Mar 28, 2020
 #1
avatar+21017 
+1

Answer for the first question.

All maxima and minima have values at cornerpoints.

The cornerpoints are (125, 225), (125, 350), and (250, 100) for values of (e, s).

At e at $6.00 and s at $12.00 per liter, the revenue produced is:

point A:  (125, 225)  =  $3450.00      with s providing $2700.00

point B:  (125, 350)  =  $4950.00      with s providing $4200.00     <---  currently the max

point C:  (250, 100)  =  $2700.00      with s providing $1200.00

 

Increasing the price of e and keeping he price of s constant will never provide an amount for point A that is

larger than the amount for point B.

So, we just need to compare the amount of point B with the amount of point C.

 

Let E be the cost per liter of e.

point B:  amount:  125E + 4200

point C: amount:   250E + 1200

 

When will these two amount be the same?   125E + 4200  =  250E + 1200

                                                                                   3000  =  125E

 --->   E = $24.00 per liter

Below $24.00 per liter, point B is the max; above $24.00 per liter, point C is the max.

 Mar 28, 2020
 #2
avatar+21017 
+1

Suggestion for the second question:

1)  Keep all the restraints (equations) for the original problem.

2)  Add these restraints:  s >= 250     e <= 200     e <= 300

3)  Graph all the original and the new equations and find their corner points.

4)  Place the values of these corner points into the revenue formula:  revenue  =  6e + 12s

     to find the maximum revenue.

 Mar 28, 2020
 #3
avatar
+1

Thank you so much @geno3141 

Guest Mar 29, 2020

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