# Maximise and minimise Z=3x-4y, subject to $\qquad$ \begin{array}{1 1}x-2y\leq 0\\-3x+y\leq 4\\x-y\leq 6\\x,y\geq 0\end{array}

Toolbox:
• Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
The objective function function given is $Z=3x-4y$ which is subjected to the constraints $x-2y\leq 0,-3x+y\leq 4,x-y\leq 6$ and $x,y\geq 0$
Now let us plot the points in the graph ,to find out the feasible region .
The line AB represents the equation $x-2y=0$,the line CD represents the equation $-3x+y=4$,the line EF represents the equation $x-y=6$
Step 2:
The feasible region bounded the above lines is OFE
The coordinates of O is the region (0,0)
The coordinates of F is the intersection of the line EF and the x-axis(6,0).
The coordinates of E is the intersection of the line EF and AB (i.e) (12,6)
Step 3:
Now let us find the optimal solutions of the objective function $Z=3x-4y$
At the points $(x,y)$ the values of the objective function subjected to $Z=3x-4y$
At $O(0,0)$ the value of the objective function $Z=3(0)-4(0)=0$
At $E(12,6)$ the value of the objective function $Z=3(12)-4(6)=12$
At $F(6,0)$ the value of the objective function $Z=3(6)-4(0)=18$
Hence the maximum value is 18.
The maximum value is 12.