SOLUTION: 3. Windies Bats produces two models of cricket bats,the Layle and the Gara. They are produced on two separate assembly lines. Producing a Layle requires 2 hours on Line I and 1 ho

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Question 1139263: 3.
Windies Bats produces two models of cricket bats,the Layle and the Gara. They are produced on two separate assembly lines. Producing a Layle requires 2 hours on Line I and 1 hour on Line II, while producing a Gara requires 1 hour on Line I and 3 hours on Line II. Sixty hours available for production on Line I and forty on Line II. The unit profit on the Layle is $150 while that on the Gara is $200.
a. Write a model to describe the information presented.
The diagram above shows the solution set shaded purple. Matching it to the information given:
b. Determine how many of each model maximizes the profit?
c. What is the maximum possible profit?

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
3. Windies Bats produces two models of cricket bats,the Layle and the Gara. They
are produced on two separate assembly lines. Producing a Layle requires 2 hours
on Line I and 1 hour on Line II, while producing a Gara requires 1 hour on Line
I and 3 hours on Line II. Sixty hours available for production on Line I and
forty on Line II. The unit profit on the Layle is $150 while that on the Gara is
$200.

.....how many of each model maximizes the profit?

Let x = the number of Layles and y = the number of Garas.

Producing a Layle requires 2 hours on Line I....., while producing a Gara
requires 1 hour on Line I..... Sixty hours available for production on Line
I.....

2x+%2B+1y+%3C=+60  (the line I inequality)

Producing a Layle requires.....1 hour on Line II, while producing a Gara
requires....3 hours on Line II.  …..forty (hours available for production) on
Line II.

1x%2B3y+%3C=+40   (the line II inequality)

The unit profit on the Layle is $150 while that on the Gara is $200.

P=150x%2B200y

a. Write a model to describe the information presented.

Maximize P=150x%2B200y
subject to constraints
2x+%2B+y+%3C=+60
x%2B3y+%3C=+40
x%3E=0, y%3E=0

The last two constraints tell us that the feasible region is
in the upper right hand part of the graph, QI

We draw the boundary line 2x+y=60, which has intercepts (30,0) and (0,60)

We draw the boundary line x+3y=40, which has intercepts (40,0) and (0,13 1/3)

We solve the system of the boundary lines by substitution or elimination
to find any points where they intersect which are vertices of the feasible region:

system%282x%2By=60%2Cx%2B3y=40%29
 
We get their point of intersection as (x,y) = (28,4)



We test the objective function at each of the four corner points:

(0,0)  P = 150x+200y = 150(0)+200(0) = 0+0 = 0
(30,0)  P = 150x+200y = 150(30)+200(0) = 4500+0 = 4500
(28,4)  P = 150x+200y = 150(28)+200(4) = 4200+890 = 5000
(0,0)  P = 150x+200y = 150(0)+200(4/30) = 0+8000/3 = 2666.67

So they should make 28 Layles and 4 Garas for a maximum profit of $5000

Edwin