The maximum value of the objective function z=4x+5y subject to constraints 2x+3y 12, 2x+y 8 and x 0 ,y 0 is

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Correct answer: C. (c) : Given inequalities are 2x+3y 12 ,2x+y 8 and x 0 ,y 0 Corresponding equations are 2x+3y=12 and2x+y=8 Now, plotting the graph of 2x+3y=12 using the table x 0 6 y 4 0 and 2x+y=8 using the table x 0 4 y 8 0 Intersection point of both lines is (3 ,2 ). For inequality 2x+3y 12 by substituting (x ,y )= (0 ,0 ) we get 0+0 12, which is true, therefore, shaded region will be towards origin. Now, for 2x+y 8 ,0+0 8, which is true. ∴ Shaded region will be towards origin. ∴ Required region is ABCO having points A (0 ,4 ), B (3 ,2 ) ,C (4 ,0 ) ,O (0 ,0 ) Max. z=4 3+5 2=12+10=22 at B (3 ,2 )

$The maximum value of the objective function z = 4 x + 5 y subject to constraints 2 x + 3 y \leq 12, 2 x + y \leq 8 and x \geq 0, y \geq 0 is$

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