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

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

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.