# Minimise $Z=13x-15y$ subject to the constraints: $x+y\leq 7,2x-3y+6\geq 0,x\geq 0,y\geq 0$.

$\begin{array}{1 1}-21 \\ -30 \\ 0 \\ \text{none of the above} \end{array}$

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:
Given : $z=13x-15y$ which is subject to the constraints $x+y\leq 7,2x-3y+6\geq 0,x\geq 0,y\geq 0$
Now let us convert the in equalities to equation,we obtain the following equation :
$x+y=7$ and $2x-3y=-6$
The line $x+y=7$ meets the $x$-axis at $(-3,0)$ and $(0,2)$
Step 2:
The point of intersection of the two lines can be obtained by solving the two equation.
$x+y=7$------(1)
$2x-3y=-6$------(2)
$\Rightarrow 2x+2y=14$
$\quad 2x-3y=-6$
_____________________
$\quad 5y=20$
$\Rightarrow y=4$
Substituting this in equation (1) we get,
$x+4=7$
$x=7-4$
$x=3$
Step 3:
Hence the point of intersection is $(3,4)$.Clearly the origin $(0,0)$ satisfies the equations.
Hence the coordinates of the vertices(corner points) are $O(0,0),A(7,0),B(3,4)$ and $C(0,2)$
Step 4:
The values of the objective function at these points are as follows :
For the point $O(0,0)$ the value of the objective function is $z=13x-15y\Rightarrow 13\times 0-15\times 0=0$
For the point $A(7,0)$ the value of the objective function is $z=13x-15y\Rightarrow 13\times 7-15\times 0=91$
For the point $B(3,4)$ the value of the objective function is $z=13x-15y\Rightarrow 13\times 3-15\times 4=-21$
For the point $C(0,2)$ the value of the objective function is $z=13x-15y\Rightarrow 13\times 0-15\times 2=-30$
Hence the minimum value is $-30$ at $(0,2)$
answered Aug 20, 2013