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We get the following conditions,x0,y0,x+y=5,x2,y4
As the maximum value of x is 2, 
So the values of x may be x = 0, 1, 2.
And as the maximum value of y is 4,
So the values of y may be y = 0, 1, 2, 3, 4.
And according to another condition x + y = 5.
This also approves the first condition because here x+y=2+3 = 5. 

That mean's if we take x = 2 and y = 3,
Then x+y=5 equation will be true.
From here, we get the maximum value of x = 2 and
The maximum value of y = 3.
The maximum value of z= 6x+2y 

z=6×22×3z=12+6z=18

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