Answer:
Maximum profit at (3,0) is $27.
Step-by-step explanation:
Let quantity of products A=x
Quantity of products B=y
Product A takes time on machine L=2 hours
Product A takes time on machine M=2 hours
Product B takes time on machineL= 4 hours
Product B takes time on machine M=3 hours
Machine L can used total time= 8hours
Machine M can used total time= 6hours
Profit on product A= $9
Profit on product B=$7
According to question
Objective function Z=[tex]9x+7y[/tex]
Constraints:
[tex]2x+3y\leq 6[/tex]
[tex]2x+4y\leq 8[/tex]
Where [tex]x\geq 0, y\geq 0[/tex]
I equation [tex]2x+3y\leq 6[/tex]
I equation in inequality change into equality we get
[tex]2x+3y=6[/tex]
Put x=0 then we get
y=2
If we put y=0 then we get
x= 3
Therefore , we get two points A (0,2) and B (3,0) and plot the graph for equation I
Now put x=0 and y=0 in I equation in inequality
Then we get [tex]0\leq 6[/tex]
Hence, this equation is true then shaded regoin is below the line .
Similarly , for II equation
First change inequality equation into equality equation
we get [tex]2x+4y=8[/tex]
Put x= 0 then we get
y=2
Put y=0 Then we get
x=4
Therefore, we get two points C(0,2)a nd D(4,0) and plot the graph for equation II
Point A and C are same
Put x=0 and y=0 in the in inequality equation II then we get
[tex]0\leq 8[/tex]
Hence, this equation is true .Therefore, the shaded region is below the line.
By graph we can see both line intersect at the points A(0,2)
The feasible region is AOBA and bounded.
To find the value of objective function on points
A (0,2), O(0,0) and B(3,0)
Put A(0,2)
Z= [tex]9\times 0+7\times 2=14[/tex]
At point O(0,0)
Z=0
At point B(3,0)
Z=[tex]9\times3+7\times0=27[/tex]
Hence maximum value of z= 27 at point B(3,0)
Therefore, the maximum profit is $27.
