Question

By graphing the system of constraints find the values of x and y that maximize the objective function, find the maximum value.

3x+y<=7
x+2y<=9
x>=0
y>=0

Maximum for P=2x+y

A. P=3
B. P=4
C. P=6
D. P=10

Answer 1

x=1, y=4

P=6

Explanation:

We are given that

`3x+y\leq 7`

`x+2y\leq 9`

`x\geq 0`

`y\geq 0`

Objective function

`P=2x+y`

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

`3x+y=7` (I equation )

`x+2y=9` (II equation)

Equation I multiply by 2 and then subtract from II equation

`-5x=-5`

`x=1`

Substitute x=1 in equation I

Then, we get

`3(1)+y=7`

`3+y=7`

`y=7-3=4`

The two equation intersect at point (1,4).

Substitute x=0 in equation I

Then , we get y=7

Substitute y=0 then

`3x=7`

`x=(7)/(3)=2.3`

The equation I cut the x- axis at point (2.3,0)and y-axis at (0,7).

Substitute x=0 in equtaion II

`2y=9`

`y=(9)/(2)=4.5`

Substitute y=0

Then, we get

`x=9`

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

`3(0)+0=0< 7`

It is true . Therefore, shaded region below the line.

Substitute x=0 and y=0 in inequality equation II

Then,
`0+2(0)=0 < 9`

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)

`P=2(0)+4.5=4.5`

At (2.3,0)

`P=2(2.3)+0=4.6`

At (1,4)

`P=2(1)+4=6`

At (0,0)

`P=2(0)+0=0`

Hence, maximum value of P is 6 at (1,4).

x=1 and y=4