Question

Minimize Z = 3x + 5y, subject to constraints are x + 3y $\ge$ 3, $x+y\ge 2andx,y\ge 0$.

Answer 1

Our problem is to minimize Z = 3x + 5y .........(i)

Subject to constraints x + 3y $\ge$ 3 .........(ii)

x + y $\ge$ 2 .........(iii)

x $\ge$ 0, y $\ge$ 0 ...(iv)

Firstly, draw the graph of the line x + 3y = 3

$\begin{array}{ccc}x& 0& 3\\ y& 1& 0\end{array}$

Putting (0, 0) in the inequality x + 3y $\ge$3, we have

$0+3\times 0\ge 3\Rightarrow 0\ge 3$ (which is false)

So, the half plane is away from the origin. Since, x, y $\ge$ 0

So, the feasible region lies in the first quadrant.

Secondly, draw the graph of the line x + y = 2

$\begin{array}{ccc}x& 0& 2\\ y& 2& 0\end{array}$

Putting (0, 0) in the inequality x + y $\ge$2, we have 0 + 0 $\ge$ 2, $\Rightarrow$ 0 $\ge$ 2 (which is false)

So, the half plane is away from the origin. It can be seen that the feasible region is unbounded.

On solving equations x + y = 2 and x + 3y = 3, we get

$x=\frac{3}{2}andy=\frac{1}{2}$

$\therefore$ Intersection point $B(\frac{3}{2},\frac{1}{2})$

The corner points of the feasible region are O(0, 0), A(3, 0), $B(\frac{3}{2},\frac{1}{2})$ and C(0, 2). The values of Z at these points are as follows :

$\begin{array}{cc}Cornerpoint& Z=3x+5y\\ O(0,0)& 0\\ A(3,0)& 9\\ B(\frac{3}{2},\frac{1}{2})& 7\to Mininum\\ C(0,2)& 10\end{array}$

As the feasible region is unbounded therefore, 7 may or may not be the minimum value of Z.

For this, we draw the graph of the inequality, $3x+5y<7$ and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with 3x + 5y < 7.

Therefore, the minimum value of Z is 7 at $(\frac{3}{2},\frac{1}{2})$.