+0

0
72
3
+90

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

Mar 28, 2020

#1
+21017
+1

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
+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
+1

Thank you so much @geno3141

Guest Mar 29, 2020