Answer
One such problem has been described below.
Work Step by Step
Sometimes, it is necessary to maximize or minimize the quantities involved in many problems to get the solution or to get the clarity of facts for taking decision. For this use of linear algebraic expressions are highly useful.
Objective function in the form of an algebraic linear expression is described in one or more variables that must be maximized or minimized according to the constraints given and need of the solution. Method that is used to do this is called linear programming method.
There are many day-to-day problems, where everyone tries to maximize the resources, time and money to get the best. In everyday life, housewife tries to manage its daily budget to meet its resources with its expenses.
Very common example is to purchase two fruits, apple and banana maximally for Rs.200 by the house wife, where apple costs Rs.180 per kg and banana costs Rs.60 per kg. The housewife has a budget of Rs.200 for the fruits and the refrigerator can store a maximum of 2 kilograms of fruits.
She must do her best for the purchase of fruits. How to purchase both the fruits to get the maximum quantity in Rs.200 is the question.
Let’s assume that she purchases x kg apple and y kg banana. Then, the objective function maximized would be:
\[z=180x+60y\]
Money constraint implies that both the fruits must cost no more than Rs.200.
This gives:
\[\begin{align}
& 180x+60y\le 200 \\
& 9x+3y\le 10
\end{align}\]
Next constraint is that the sum of the weight of both the fruits must not be more than 2kgs. This implies:
\[x+y\le 2\]
Also, as both the variables denote weights of the two kinds of fruits, non-negative constraints would also be applicable in this situation:
\[\begin{align}
& x\ge 0 \\
& y\ge 0
\end{align}\].
Now, try to maximize the value of objective function\[z=180x+60y\] according to the above constraints.
Now to test for the shaded region, consider the test point \[\left( 0,0 \right)\]. Substitute this point in both the inequalities to see where the shaded region lies.
For \[9x+3y\le 10\],
\[\begin{align}
& 9\left( 0 \right)+3\left( 0 \right)\le 10 \\
& 0\le 10
\end{align}\]
As, the test point satisfies the inequality, the shaded region would be towards the test point.
For \[x+y\le 2\],
\[\begin{align}
& \left( 0 \right)+\left( 0 \right)\le 2 \\
& 0\le 2
\end{align}\]
As, the test point satisfies the inequality, the shaded region would be towards the test point.
The shaded region shows just one intersection point.
This point can be computed as:
\[9x+3y-10=x+y-2\]
This can be solved by multiplying the equation on the right-hand side by 3 and then subtracting it from the equation on the left-hand side.
\[\begin{align}
& 9x+3y-10-3\left( x+y-2 \right)=0 \\
& 9x+3y-10-3x-3y+6=0 \\
& 6x=4 \\
& x=\frac{2}{3}
\end{align}\]
Substitute back the value of x to obtain:
\[\begin{align}
& x+y=2 \\
& \frac{2}{3}+y=2 \\
& y=2-\frac{2}{3} \\
& =\frac{4}{3}
\end{align}\]
This implies that the optimum value for the provided linear programing problem would be \[\left( \frac{2}{3},\frac{4}{3} \right)\].
