This is a complicated problem!
The costs of the fertilizers are not given; I will assume that the profits per ton include the costs of the fertilizers when applied in the suggested amounts and proportions. We do need to take into account the additional costs incurred if some or all of fertilizer A is replaced by improved fertilizer B for the tomatoes. These costs consist of the direct cost of doing the improving and the indirect cost of the resulting decrease in yield for the tomatoes.
I will assume that supplement should have been replacement.
In this question pound is a unit of currency, not weight; this problem probably originated in the United Kingdom, though a few other countries also call their primary unit of currency a pound. I will pretend that the problem was stated in dollars to avoid any confusion over the word pound.
t = acres of tomatoes
p = acres of pumpkins
a = tons of fertilizer A
b = tons of fertilizer B
i = acres of tomatoes fertilized using improved fertilizer B
t + p = 200 so p = 200 - t
Objective function: maximize profit, which is given by
(5x0.9)(50)i + (5)(50)(t-i) + (3)(100)(200 – t) – (45)(0.4)i
= 225i + 250t – 250i + 60000 – 300t – 18i
= 60,000 – 50t – 43i
Explanation: Before simplifying we have (5x0.9)(50)i + (5)(50)(t-i) + (3)(100)(200 – t) – (45)(0.4)i. Profit is acres x tons produced per acre x dollars per ton, with an adjustment at the end for the direct extra cost of improving fertilizer. The first term takes into account the indirect cost of using improved fertilizer B, namely a 10% reduction in yield of tomatoes on the acres for which it is used. The second term is the profit from the acres of tomatoes on which improved fertilizer B is not used, for which there is no reduction in yield. The third term is the profit from growing pumpkins; 200 – t is the acres devoted to pumpkins. The fourth term is the direct cost of improving fertilizer B, namely cost per ton x tons per acre x acres of tomatoes on which it is used, for which it is the only fertilizer used.
a. i < t so t – i > 0
Only tomatoes can be fertilized using improved B, so the acres fertilized using it cannot exceed the total acres planted in tomatoes.
b. t < 200
Acres of tomatoes cannot exceed the total available acres.
c. a < 30:
a = (0.4x0.4)(t-i) + (0.5x0.55)(200 – t) = 0.16t – 0.16i + 55 – 0.275t = 55 – 0.115t – 0.16i
substitute that for a in inequality: 55 – 0.115t – 0.16i < 30
so -0.115t – 0.161i < -25
so 0.115t + 0.161i > 25
Explanation: Amount of fertilizer A used cannot exceed 30 tons; amount used = (tons per acre)(fraction of fertilizer that is A)(acres), keeping in mind that acres of tomatoes fertilized using improved fertilizer B are fertilized 100% with that, and use no fertilizer A, and 200 – t is acres of pumpkins.
d. b < 100:
b = (0.4x0.6)t + (0.4x0.4)i + (0.5x0.45)(200 – t) = 0.24t + 0.16i + 45 – 0.225t = 45 + 0.015t + 0.16i
substitute that for b in inequality: 45 + 0.015t + 0.16i < 100
so 0.015t + 0.16i < 55
Expanation: Similar to c., but fertilizer B is used at a rate of 60% of total fertilizer for tomatoes plus the remaining 40% for the acres of tomatoes fertilized using improved B, for which it is the only fertilizer used.
You might wonder why one would possibly want to use improved fertilizer B when it reduces yield and costs more. The answer lies in the limited amount of fertilizer A available, and the fact that improved B can substitute for A for tomatoes, thus conserving the limited amount of A for use on the more profitable pumpkins. Why more profitable? The 40% smaller tons per acre yield is more than made up for by the 100% higher tons per acre profit. By contrast, the constraint on B will turn out to not matter; if one grew only tomatoes and used improved B on all of it, there would still be enough B: 200 acres x 0.4 tons of fertilizer per acre would be only 80 tons, which is less than the maximum for B of 100 tons.
Since we are down to two decision variables, we can use the graphing method to solve the
problem. I graphed t on the x axis and i on the y axis. For graphing purposes, we need to restate
each constraint solved for i (except for b, which is a vertical line) and as an equality rather than
a. i = t
b. t = 200
c. i = (25 – 0.115t)/0.16 = 156.25 – 0.71875t
d. i = (55 – 0.015t)/0.16 = 343.75 – 0.09375t
Unfortunately, this system won’t let me include an image in my answer, so I’ll tell you what it looks like. The line for a. goes up as it goes to the right, the line for b. is vertical at t = 200, and the line for c. goes downward as it goes to the right. These combine to form a triangular feasible area with vertices at the intersections of a. with b., a. with c., and b. with c. The line for d. goes downward to the right but well above the feasible triangle; it does not play a part in our solution, as expected.
The optimum solution will be at one of the vertices of the feasible triangle; we need to find the values of t and i at each vertex and plug them into the objective function to see which vertex is best.
a meets b:
i = t = 200 acres; p = 200 – t = 0
objective function = 60,000 – 50x200 – 43x200 = 41,400 dollars profit
a meets c:
i = t = 156.25 – 0.71875t
t + 0.71875t = 1.71875t = 156.25
t = 156.25/1.71875 = 90.90909 acres planted in tomatoes = i so tomatoes fertilized entirely with improved fertilizer B; p = 200 – t = 200 - 90.90909 = 109.09091 acres planted in pumpkins
objective function = 60,000 – 50 x 90.90909 – 43 x 90.90909 = 51,545 dollars profit
b meets c:
t = 200
i = 156.25 – 0.71875x200 = 12.5
objective function = 60,000 – 50x200 – 43x12.5 = 49,463 dollars profit
So the solution is the values at the vertex where a. meets c.: 90.90909 acres tomatoes, all of those acres fertilized using improved fertilizer B, and 200 – 90.90909 = 109.09091 acres pumpkins.
How much fertilizer A is used? 109.09091 x 0.5 x 0.55 = 30.00000 tons (limit was 30).
In the above calculations I showed far more decimal places than would be called for based on
significant figures because I wanted to show that the amount of fertilizer A used worked out to
be exactly at its limit.