Question

Maximize $Z=5x+3y$, subject to constraints $3x+5y\le 15,5x+2y\le 10,x\ge 0$ and $y\ge 0$.

Answer 1

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

Subject to constraints $3x+5y\le 15$ .......(ii)

$5x+2y\le 10$ ..........(iii)

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

Firstly, draw the graph of the line $3x+5y=15$$\begin{array}{ccc}x& 0& 5\\ y& 3& 0\end{array}$
Putting $(0,0)$ in the inequality $3x+5y\le 15$, we have
$(3\times 0)+(5\times 0)\le 15\Rightarrow 0\le 15$ (which is true)
So, the half plane is towards the origin. Since, $x,y\ge 0$
So, the feasible region lies in the first quadrant.

Secondly, draw the graph of the line $5x+2y=10$
$\begin{array}{ccc}x& 0& 2\\ y& 5& 0\end{array}$
Putting $(0,0)$ in the inequality $5x+2y\le 10$, we have $(5\times 0)+(2\times 0)\le 10$

$\Rightarrow 0\le 10$ (which is true)

So, the half plane is towards the origin. On solving given equations $3x+5y=15$ and $5x+2y=10$,

we get $x=\frac{20}{19},y=\frac{45}{19}$

Coordinates of point $B$ is( $\frac{20}{19},\frac{45}{19}$).

Feasible region is $OABCO$.

The corner points of the feasible region are $O(0,0),A(2,0),B(\frac{20}{19},\frac{45}{19})$ and $C(0,3)$ The values of $Z$ at these points are as follows :
$\begin{array}{cc}Cornerpoint& Z=5x+3y\\ O(0,0)& 0\\ A(2,0)& 10\\ C(0,3)& 9\\ B(\frac{20}{19},\frac{45}{19})& \frac{235}{19}\to Mininum\end{array}$

Therefore, the maximum value of $Z$ is $\frac{235}{19}$ at the point $B(\frac{20}{19},\frac{45}{19})$.