Hey yall I really need some help; Image attached below - 50 points
1 answer:
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Answer: It should be used 2 for type-A and 3 for type-B to minimize the cost.
Step-by-step explanation: As it is stipulated, <u>x</u> relates to type-A and y to type-B.
Type-A has 60 deluxe cabins and B has 80. It is needed a minimum of 360 deluxe cabins, so:
60x + 80y ≤ 360
For the standard cabin, there are in A 160 and in B 120. The need is for 680, so:
160x + 120y ≤ 680
To calculate how many of each type you need:
60x + 80y ≤ 360
160x + 120y ≤ 680
Isolating x from the first equation:
x =
Substituing x into the second equation:
160() + 120y = 680
-3200y+1800y = 10200 - 14400
1400y = 4200
y = 3
With y, find x:
x =
x =
x = 2
To determine the cost:
cost = 42,000x + 51,000y
cost = 42000.2 + 51000.3
cost = 161400
To keep it in a minimun cost, it is needed 2 vessels of Type-A and 3 vessels of Type-B, to a cost of $161400
Answer:
13
Step-by-step explanation:
HCF = Product of two numbers ÷ LCM of these numbers
HCF =
= 13