Answer:
x=1, y=4
P=6
Step-by-step explanation:
We are given that
[tex]3x+y\leq 7[/tex]
[tex]x+2y\leq 9[/tex]
[tex]x\geq 0[/tex]
[tex]y\geq 0[/tex]
Objective function
[tex]P=2x+y[/tex]
We have to values of x and y that maximize the P and maximum value of P .
First we change inequality equation into equality equation
[tex]3x+y=7[/tex] (I equation )
[tex]x+2y=9[/tex] (II equation)
Equation I multiply by 2 and then subtract from II equation
[tex]-5x=-5[/tex]
[tex]x=1[/tex]
Substitute x=1 in equation I
Then, we get
[tex]3(1)+y=7[/tex]
[tex]3+y=7[/tex]
[tex]y=7-3=4[/tex]
The two equation intersect at point (1,4).
Substitute x=0 in equation I
Then , we get y=7
Substitute y=0 then
[tex]3x=7[/tex]
[tex]x=\frac{7}{3}=2.3[/tex]
The equation I cut the x- axis at point (2.3,0)and y-axis at (0,7).
Substitute x=0 in equtaion II
[tex]2y=9[/tex]
[tex]y=\frac{9}{2}=4.5[/tex]
Substitute y=0
Then, we get
[tex]x=9[/tex]
Therefore, the equation II cut the x- axis at point (9,0) and y axis at point (0,4.5).
Substitute x=0 and y=0 in inequality Equation I
[tex]3(0)+0=0< 7[/tex]
It is true . Therefore, shaded region below the line.
Substitute x=0 and y=0 in inequality equation II
Then, [tex]0+2(0)=0 < 9[/tex]
It is true. Therefore, the shaded region below the line.
The feasible region is bounded.The feasible region bounded by (0,0),(0,4.5),(2.3,0) and (1,4).
At (0,4.5)
[tex]P=2(0)+4.5=4.5[/tex]
At (2.3,0)
[tex]P=2(2.3)+0=4.6[/tex]
At (1,4)
[tex]P=2(1)+4=6[/tex]
At (0,0)
[tex]P=2(0)+0=0[/tex]
Hence, maximum value of P is 6 at (1,4).
x=1 and y=4
