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