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The minimum value of z=3x+2yz = 3x + 2y subjected to the constraints 2x+y≥7,x+2y≥8,x,y≥02x + y \geq 7, x + 2y \geq 8, x, y \geq 0 is

Solution

✅ Correct Option: 3

The objective is to find the minimum value of z=3x+2yz = 3x + 2y subject to:

  • 2x+y≥72x + y \geq 7
  • x+2y≥8x + 2y \geq 8
  • x,y≥0x, y \geq 0

In linear programming, minimum or maximum values occur at corner points (vertices) of the feasible region.


To find the corner point, solve the constraint equations simultaneously:

2x+y=72x + y = 7 ... (equation 1)

x+2y=8x + 2y = 8 ... (equation 2)

From equation 1:

y=7−2xy = 7 - 2x

Substitute into equation 2:

x+2(7−2x)=8x + 2(7 - 2x) = 8

x+14−4x=8x + 14 - 4x = 8

−3x=−6-3x = -6

x=2x = 2

Find yy:

y=7−2(2)y = 7 - 2(2)

y=7−4y = 7 - 4

y=3y = 3

The corner point is (2,3)(2, 3).


Substitute (2,3)(2, 3) into z=3x+2yz = 3x + 2y:

z=3(2)+2(3)z = 3(2) + 2(3)

z=6+6z = 6 + 6

z=12z = 12

Therefore, the minimum value is 1212.

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