Given:
The objective function, [tex]C=2x+3y[/tex],
Constraints,
[tex]2x+3y\leq 24[/tex]
[tex]3x+y\leq 15[/tex]
[tex]x\geq 2[/tex]
[tex]y\geq 3[/tex]
To find:
The minimum value of the given objective function.
Solution:
We have,
Objective function, [tex]C=2x+3y[/tex] ....(i)
The related equations of given constraints are
[tex]2x+3y=24[/tex] ...(ii)
[tex]3x+y=15[/tex] ...(iii)
[tex]x=2[/tex], it is a vertical line parallel to y-axis and 2 units left from y-axis.
[tex]y=3[/tex], it is a horizontal line parallel to x-axis and 4 units above from y-axis.
Table of value for (i),
x : 0 12
y : 8 0
Table of value for (ii),
x : 0 5
y : 15 0
Plot these points on a coordinate plane and draw these 4 related lines.
The sign of inequality of [tex]2x+3y\leq 24[/tex] and [tex]3x+y\leq 15[/tex] is ≤, it means the related lines are solid lines and the shaded region lie below the related line.
For [tex]x\geq 2[/tex], left side of the line [tex]x=2[/tex] is shaded.
For [tex]y\geq 3[/tex], shaded region is above the line [tex]y=3[/tex].
From the below graph it is clear that the vertices of the feasible (common shaded region) are (2,3), (4,3), (3,6) and (2,6.667).
Points [tex]C=2x+3y[/tex]
(2,3) [tex]C=2(2)+3(3)=4+9=13[/tex] (Minimum)
(4,3) [tex]C=2(4)+3(3)=8+9=17[/tex]
(3,6) [tex]C=2(3)+3(6)=6+18=24[/tex]
(2,6.667) [tex]C=2(2)+3(6.667)=4+20=24[/tex]
Therefore, the objective function is minimum at point (2,3).
