SOLUTION: I've tried and am completely lost on this.. Any help would be highly appreciated! A company wishes to produce two types of souvenirs: Type A and Type B. Each Type A souvenir will r

Question 1201272: I've tried and am completely lost on this.. Any help would be highly appreciated! A company wishes to produce two types of souvenirs: Type A and Type B. Each Type A souvenir will result in a profit of $1.00, and each Type B souvenir will result in a profit of $1.40. To manufacture a Type A souvenir requires 2 minutes on Machine I and 1 minute on Machine II. A Type B souvenir requires 1 minute on Machine I and 3 minutes on Machine II. There are 3 hours available on Machine I and 5 hours available on Machine II.
(a) For a meaningful solution, the time available on Machine II must lie between
and min. (Enter your answers from smallest to largest.)
(b) If the time available on Machine II is changed from 300 min to (300 + k)
min, with no change in the maximum capacity for Machine I, then Ace Novelty's profit is maximized by producing
Type A souvenirs and
Type B souvenirs, where
? ≤ k ≤?

(c) Find the shadow price for Resource 2 (associated with constraint 2). (Round your answer to the nearest cent.)
Answer by carrollmary87(4) (Show Source): You can put this solution on YOUR website!
I can do A and B, so I hope that helps.
First, A is a system of equations for the two machines. Also known as linear programming. The machine I is 2A + 1B = 180 min. and Machine II is 1A + 3B = 300, so using either matrix, substitution, or elimination.
2(1A + 3B = 300) is 2A + 6B = 600 subtract 2A + 1B = 180
2A + 6B = 600
-2A - 1B = -180
5B = 420
B = 84, then substitute B into one of the equations. 1A + 3(84) = 300
A = 48
Profit then for the company is 1(48) + 1.4(84) = 165.60
B) is asking if you increase the minutes for the maximum profit. Max profit would be no A and all B. because you make more money on B.
Since the first restraint for B is machine 1: 2(0) + B = 180, then you can only make 180 B parts because of machine 1's limits.
Machine 2 is 1(0) + 3B = 300 + k where 0 + 3(180) = 300 + 240 (540 min total)
so the range for k is
1 < = k < = 240
If I understand C correctly, the answer is 1.4 (180) = $252
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