Question

Solve using graphical method:
Maximize: $Z=4x+5y$
Subject to:
$2x+5y\le 25$
$6x+5y\le 45$
$49$
$x\ge 0,y\ge 0$

Answer 1

To solve the above linear programming model using the graphical method, we shall turn each constraints inequality to equation and set each variable equal to zero (0) to obtain two (2) coordinate points for each equation (i.e using double intercept form). Having obtained all the coordinate points, we shall determine the range of our variables which enables us to know the appropriate scale to use for our graph. Thereafter, we shall draw the graph and join all the coordinate points with required straight line.
$2x+5y=25$ (Constraint 1)
When $x=0,y=5$ and when $y=0,x=12.5$
$6x+5y=45$ (Constraint 2)
When $x=0,y=9$ and
when $y=0,x=7.5$
Minimum value of $x$ is $x=0$
Maximum value of $x$ is $x=12.5$
Range of $x$ is $0\le x\le 12.5$
Minimum value of $y$ is $y=0$
Maximum value of $y$ is $y=9$
The constrains give a set of feasible solutions as graphed. To solve the linear programming problem, we must now find the feasible solution that makes the objective function as large as possible. Some possible solution are listed below

Feasible solution (A point in the solution set of the system)	Objective function $Z=4x+5y$
$(2,3)$	$4\left(2\right)+5\left(3\right)=8+15=23$
$(4,2)$	$4\left(4\right)+5\left(2\right)=16+10=26$
$(5,1)$	$4\left(5\right)+5\left(1\right)=20+5=25$
$(7,0)$	$4\left(7\right)+5\left(0\right)=28+0=28$
$(0,5)$	$4\left(0\right)+5\left(5\right)=0+25=25$

In this list, the point that makes the objective function the largest is $(7,0)$. But, is this the largest for all feasible solutions? How about $(6,1)$? or $(5,3)$? IT turns out that $(5,3)$ provide the maximum value; $4\left(5\right)+5\left(3\right)=20+15=35$
Hence, the maximum profit at point $(5,3)$ and it is the objective functions which have optimal values