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lbvjy [14]
3 years ago
7

Your coffee bar has two kinds of drink One cup of the first one costs 9 grams of juice , 4 grams of coffee and 3 grams of sugar

The second one costs 4 grams of juice, 5 grams of coffee and 10 grams of sugar. The budget at most 3.6 kilograms of Julce, 2.0 kilograms of coffee and 3.0 kilograms of sugar every day. You get $1.2 if you sell one cup of the first kind of drink and $0.7 of the second one. aWhat is the maximal profit $ b) How many drinks should be sold to get a maximal profit ? Sales of the first one = cups Sales of the second one = cups
Mathematics
1 answer:
VMariaS [17]3 years ago
6 0

Answer:

How many drinks should be sold to get a maximal profit? 468

Sales of the first one = 345 cups

Sales of the second one = 123 cups

Step-by-step explanation:

maximize 1.2F + 0.7S

where:

F = first type of drink

S = second type of drink

constraints:

sugar ⇒ 3F + 10S ≤  3000

juice ⇒ 9F + 4S ≤  3600

coffee ⇒ 4F + 5S ≤  2000

using solver the maximum profit is $500.10

and the optimal solution is 345F + 123S

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Answer:

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Step-by-step explanation:

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3 years ago
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Eduardwww [97]

Answer:

1. 53

2. 20

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Step-by-step explanation:

Since your trying to find the hypotenuse you equation is...

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So after you square root 2809 you get 53.

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3 years ago
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Nutka1998 [239]

Answer:

The answer to the question is C

4 0
3 years ago
A store charges $32 to print 80 greeting cards and $0.33 for each additional greeting card.
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