+90 This a link to the question, it includes a graph as well.
https://free-picload.com/image/qtfKRh
+23246 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.
+23246 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.
