Algebra 2 Linear Programming
posted by Chauncey .
I think this is an easy question but i just wanted make sure i have the right answer
Could you check my work
Some students make and sell jewelry in their spare time. Every week they have an avaiable 10,000 beads of various colors and sizes. They only have 20 hours to work on the jewelry. Each necklace takes 30 minutes to make and each bracelet takes 20 minutes to make. Only 50 beads are placed on each necklace,and 200 beads on each bracelet(yes the bracelet has more beads then the necklace, not a typo). The profit on every necklace is $3.50 and the profit on every bracelet is $2.50. The students want to make as much money as possible.
Define variable: x= necklace
y= bracelets
Write constraints
Write objective function for profit: i think it is P=3.5X+2.5Y
State the vertices in a chart....if i could just get the vertices that would be nice....i think it is (50,30)
Evaluate the profit for each point, but i think there is only one point in this problem
What is the maximum profit
i think it is 3,400
thanks for everything

Algebra 2 Linear Programming 
Damon
agree with P = 3.5 x + 2.5 y
But constraints
Number of beads :
N = 50 x + 200 y must be </= 10,000
200 y = 50 x + 10,000
so on your graph
y =  .25 x + 50
Number of hours
Hr = .5 x + .333 y must be </= 20
.333 y = .5 x + 20
so on your graph
y = 1.5 x + 60
so graph it
The number of beads line hits the y axis at 50 and the x axis at 200
The number of hours line hits the y axis at 60 and the x axis at 40
the two lines cross at
x = 8, y = 48 (8,48)
SO VERTICES
(0,0) of course
(0,50) all bracelets
(8,48)
(40,0) all necklaces
NOW do your profit at each point (4 including the trivial one at origin) 
Algebra 2 Linear Programming 
Damon
Profit at (0,0) = 0 of course
Profit at (0,50) = 3.5(0)+2.5(50) =125
Profit at (8,48) = 3.5(8)+2.5(48) =148
Profit at (40,0) = 3.5(40)+2.5(0) =140
so make 8 necklaces and 48 bracelets
Do you agree? 
Algebra 2 Linear Programming 
Chauncey
that sounds right i guess i should check my work again
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