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
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)
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?
that sounds right i guess i should check my work again
To solve this problem, we need to find the maximum profit by maximizing the number of necklaces (x) and bracelets (y) while staying within the given constraints.
First, let's write the constraints:
1. The number of beads used for necklaces (50 beads) times the number of necklaces (x) plus the number of beads used for bracelets (200 beads) times the number of bracelets (y) should not exceed the available beads (10,000 beads).
This can be written as: 50x + 200y ≤ 10,000.
2. The total time taken to make necklaces (30 minutes each) times the number of necklaces (x) plus the total time taken to make bracelets (20 minutes each) times the number of bracelets (y) should not exceed the available time (20 hours = 1200 minutes).
This can be written as: 30x + 20y ≤ 1200.
3. The number of necklaces (x) and bracelets (y) should be greater than or equal to zero.
This can be written as: x ≥ 0 and y ≥ 0.
Next, let's write the objective function for profit, where P represents the total profit:
P = 3.5x + 2.5y.
Now, we can find the vertices by solving the system of equations formed by the constraints:
50x + 200y ≤ 10,000
30x + 20y ≤ 1200
x ≥ 0
y ≥ 0
Solving these equations will give us the vertices or the points of intersection.
The maximum profit will be achieved at one of these vertices. We need to evaluate the profit (P) at each vertex to determine the maximum profit.
The coordinates of the vertices are found by solving the system of equations.
After solving the system, we find that the vertices are:
(0, 0), (6, 45), (40, 30), (200, 0).
Now, we can calculate the profit at each vertex using the objective function:
P(0, 0) = 3.5(0) + 2.5(0) = 0
P(6, 45) = 3.5(6) + 2.5(45) = 21 + 112.5 = 133.5
P(40, 30) = 3.5(40) + 2.5(30) = 140 + 75 = 215
P(200, 0) = 3.5(200) + 2.5(0) = 700
From these calculations, it is clear that the maximum profit is $215, which occurs at the vertex (40, 30).
Therefore, the maximum profit the students can make is $215.
I hope this helps! Let me know if you have any further questions.