Answer:
Rachel should have 1875 Family Thrillseekers and 3500 Classy Cruisers assembled for a maximum profit of $24,150,000.
Explanation:
As the question is not complete, by searching the question statement, a similar question is found so that complete data is used so
Let the number of Classy cruisers be X
Let the number of Family Thrillseeker is Y
so
The profit on classy cruisers is given as $5400.(from complete question)
The profit on Family Thrillseeker is as given $2800.
The maximizing function is
[tex]Z=\$5400X+\$2800Y[/tex]
The constraints are as follows
The first constraint is as The plant has a capacity of 48,000 labor-hours per month. It takes 6 labor-hours to assemble one Family Thrillseeker and 10.5 labor-hours to assemble one Classy Cruiser. given as
10.5X+6Y[tex]\leq[/tex] 48000
The second constraint is as Rachel will only be able to obtain 20,000 doors (10,000 left and 10,000 right) from the door supplier, the Family Thrillseeker(2 Left & 2 Right) and the Classy Cruiser(1 Left and 1 Right) use the same door part. Given as
2X+4Y[tex]\leq[/tex] 20000
The third constraint is as A company forecast suggests that the demand for the Classy Cruiser is limited to 3500 cars, and there is no limit on the demand of the Family Thrillseeker. Given as
X[tex]\leq[/tex] 3500
Using the plots of constraints corresponding equations and solving for the corner points as given below
P1(0,0)
P2(0,5000)
P3(2400,3800)
P4(3500,1875)
P5(3500,0)
For each of the above values, the value of function Z is calculated and maximized accordingly
[tex]Z(0,0)=\$5400(0)+\$2800(0) =0[/tex]
[tex]Z(0,5000)=\$5400(0)+\$2800(5000) =14000000[/tex]
[tex]Z(2400,3800)=\$5400(2400)+\$2800(3800) =23600000[/tex]
[tex]Z(3500,1875)=\$5400(3500)+\$2800(1875) =24150000[/tex]
[tex]Z(3500,0)=\$5400(3500)+\$2800(0) =18900000[/tex]
In this case Rachel should have 1875 Family Thrillseekers and 3500 Classy Cruisers assembled for a maximum profit of $24,150,000.
