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
Answer:
(1, 0) and (3, 0)
Step-by-step explanation:
y=2x²-8x+6
x- intercept when y=0
- 2x²-8x+6=0
- x²-4x+4=1
- (x-2)²=1
- x-2=1 ⇒ x= 3
- x-2= -1 ⇒ x= 1
