Objective function Z=3x+6yFind the value at each corner of the graph
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Given: An objective function
[tex]z=3x+6y[/tex]with constraints-
[tex]\begin{gathered} \\ \begin{cases}{x\ge0,y\ge0} \\ {2x+y\leq12} \\ {x+y\ge6}\end{cases} \end{gathered}[/tex]Required: To graph the linear inequalities representing the constraints and determine the objective function's value at each corner.
Explanation: The inequalities can be graphed by considering them as equations and then determining the shaded region by less than or greater than symbol.
The equation for the first inequality is-
[tex]2x+y=12[/tex]This represents a straight line passing through points (6,0) and (0,12).
The shaded region will be below this line as the inequality is-
[tex]2x+y\leq12[/tex]Similarly, the inequality-
[tex]x+y\ge6[/tex]Represents a shaded region above the line x+y=6.
The inequalities-
[tex]x\ge0,y\ge0[/tex]Represents the positive values of x and y. Hence we need to determine the graph in the first quadrant.
The graph of the inequalities is-
The graph in blue represents the inequality-
[tex]2x+y\leq12[/tex]While the graph in green represents the inequality-
[tex]x+y\ge6[/tex]The corner points of the common shaded area are A(0,6), B(0,12), and C(6,0).
Now the value of the objective function at these points is-
a) At A(0,6)
[tex]\begin{gathered} z=3(0)+6(6) \\ =36 \end{gathered}[/tex]b) At B(0,12)
[tex]\begin{gathered} z=3(0)+6(12) \\ =72 \end{gathered}[/tex]c) At C(6,0)
[tex]\begin{gathered} z=3(6)+6(0) \\ =18 \end{gathered}[/tex]Final Answer: a) The graph is drawn.
b) 36,72,18