Question

Maximize z=8x+9y subject to contsraints give as 2x+3y<6 ,3x-2y<6,y<1,x,y>0.

Answer 1

Answer:

( $\frac{30}{13}$, $\frac{6}{13}$ ) maximizes our given function

Step-by-step explanation:

Given function which need to be maximized

z = 8x + 9y ... (A)

Given inequalities

2x + 3y < 6 .... (i)

3x -2y <6 ...... (ii)

y<1, y>0

The corresponding equations of equation 1 and two are given as

2x + 3y = 6 .... (iii)

3x -2y = 6 ...... (iv)

Lets us find the point where these two lines cross each other, and that point will maximize our given function

comparing equation (iii) and (iv)

2x + 3y = 3x - 2y

Rearranging

3x - 2x = 2y + 3y

x = 5y ......(v)

Now plug x = 5y into equation (iv)

3(5y) - 2y = 6

15y - 2y = 6

13y = 6

y = $\frac{6}{13}$

Now put y = $\frac{6}{13}$ into equation (v)

x = 5($\frac{6}{13}$)

x = ($\frac{30}{13}$)

Now Substitute x = ($\frac{30}{13}$) and y = $\frac{6}{13}$  into equation (A)

z  =8($\frac{30}{13}$) + 9($\frac{6}{13}$)

z = ($\frac{240}{13}$) + ($\frac{54}{13}$)

z = $\frac{240 +54}{13}$

z = $\frac{294}{13}$

So ( $\frac{30}{13}$, $\frac{6}{13}$ ) maximizes our given function.