Question 1139263
<pre>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 + 1y <= 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+3y <= 40}}}   (the line II inequality)

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

{{{P=150x+200y}}}

a. Write a model to describe the information presented.

Maximize {{{P=150x+200y}}}
subject to constraints
{{{2x + y <= 60}}}
{{{x+3y <= 40}}}
{{{x>=0}}}, {{{y>=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(2x+y=60,x+3y=40)}}}
 
We get their point of intersection as (x,y) = (28,4)

{{{drawing(275,400,-5,50,-5,75,locate(28,6.4,"(28,4)"),
locate(-2,0,"(0,0)"), locate(26,0,"(30,0)"),locate(0,17,"(0,40/3)"),
locate(2,5,matrix(1,2,FEASIBLE,REGION)),
line(0,0,0,70),line(0,0,70,0), line(30,0,0,60),line(40,0,0,40/3) )}}}

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</pre>