Answer a: For L1 to be best:
Let the quantity = x
Total cost of L1 <= Total cost of L2 and L3
Total cost = fixed cost + quantity*variable cost
L1 = 13515 + 2.3*x
L2 = 8605 + 3.1*x
L3 = 6388 + 4.1*x
13515 + 2.3*x <= 8605 + 3.1*x
0.8*x >= 4910
x >= 6137.5
13515 + 2.3*x <= 6388 + 4.1*x
1.8*x >= 7127
x >= 3959.44
Q >= 6138 (rounded to nearest whole, if not required, leave accordingly)
Answer b: For alternative 2 to be best:
L1 = 13515 + 2.3*x
L2 = 8605 + 3.1*x
L3 = 6388 + 4.1*x
8605 + 3.1*x <= 13515 + 2.3*x
0.8*x <= 4910
x <= 6137.5
8605 + 3.1*x <= 6388 + 4.1*x
1*x >= 2217
x >= 2217
2217 <= Q <= 6138 (rounded to nearest whole, if not required, leave accordingly)
Answer c: For L3 to be best
L1 = 13515 + 2.3*x
L2 = 8605 + 3.1*x
L3 = 6388 + 4.1*x
6388 + 4.1*x <= 8605 + 3.1*x
1*x <= 2217
x <= 2217
6388 + 4.1*x <= 13515 + 2.3*x
1.8*x <= 7127
x <= 3959.44
Q <= 2217
Answer d: x = 1505
since, x =1505 comes in the range of L3,
The Alternative would yield the lowest total cost for an expected annual volume of 1505 boats is location L3.
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