Solve the LPP graphically

Minimize & Maximize

$z=3 x+9 y \text { (objective) }$

Subject to constraints

$x+3 y \leqslant 60 \text { - (i) }$

$x+y \geqslant 10 \text { - (ii) }$

$x \leqslant y \text {- (iii) }$

$x \geqslant 0, y \geqslant 0 \text { - (iv) }$

Solve: Associated Equation

$\begin{aligned} & x+3 y=60 \text { - (i) } \\ & x+y=10 \text { - (ii) } \\ & x=y \text { - (iii) } \end{aligned}$

from (i) $x+3 y=60$

put $y=0 \Rightarrow x=60$

$x=0 \Rightarrow y=20$

$\therefore$ points are (60, 0) & (0,20)

from (ii)

$x+y =10$

$\text { put } y =0 \Rightarrow x=10$

$x =0 \Rightarrow y=10$

$\therefore$ points an (10,0) &(0,10)

from (iii) $\quad \quad x=y$

$y =0 \Rightarrow x=0$

$x =1 \Rightarrow y=1$

$\therefore$ points are (0,0) &(1,1)

origin test

$0+3 \times 0=0<60(T)$

$0+0=0>10(F)$ $~ ~$ $0=0(T)$

$\text { for } x=y$ $~ ~$ $(1,2)$

$1<2$

(10,20) $\quad$ (T)

10<20 (T)

from (i) & (iii)

$x+3 y=60$

$\Rightarrow x+3 x=60$

$(x=y)$

$\Rightarrow \quad 4 x=60$

$\Rightarrow \quad x=15$

$\therefore y=15$

$B(15,15)$

from (ii) & (iii)

$x+y=10$

$\Rightarrow x+x=10$

$\quad(x=y)$

$\Rightarrow 2 x=10$

$\Rightarrow x=5$

$\therefore y=5$

$\therefore A(5,5)$

$\therefore$ Corner point of feasible region $A B C D$ are

$A(5,5), B(15,15),$

$C(0,20), D(0,10)$

value of $Z=3 x+9 y$ at Corner points are

$Z_A=3 \times 5+9 \times 5=60 \quad \text { (smallest) }$

$Z_B=3 \times 15+9 \times 15=45+135=180 \text { (largest) }$

$Z_C=3 \times 0+9 \times 20=0+180=180 \text { (largest) }$

$Z_D=3 \times 0+9 \times 10=0+90=90$

Since feasible region is bounded region

$\therefore Z_{\text {min. }}=60 \text { at } A(5,5)$

Since largest value at two points B and C.

$\therefore Z_{\text {max }}=180$ lies on the line $B C$.

i.e. Problem has multiple optimal solution.

Solve LPP graphically

Minimize

$Z=5 x+3 y$

Subject to

$x+y=6$

$x \leqslant 4$

$y \leqslant 5$

$x \geqslant 0, y \geqslant 0$

Solution: : Associated equation for given constraints are

$x+y=6 \text {\ldots (i) }$

$x=4 \text {\ldots (ii) }$

$y=5 \text { \ldots (iii) }$

from (i) $x+y=6$

$\text { put } y =0 \Rightarrow x=6$

$x =0 \Rightarrow y=6$

$\therefore$ points are (6,0) & (0,6)

from (ii) $x=4$ is a line parallel to $y$-axis intersect- $x$-axis at- $(4,0)$

from (iii) $y=5$ is a line parallel to $x$-axis intersect $y$ axis at $(0,5)$

feasible region is line AB with corner points

A(4,2) & B(1,5)

$\therefore Z_A=5 x+3 y$

$= 5 \times 4 + 3 \times 2$

$= 26$

Solve the LPP graphically

$\text { Minimize } Z=3 x+5 y$

$\text { Subject to }$

$x+2 y \geqslant 10$

$x+y \geqslant 6$

$3 x+y \geqslant 8$

$x \geqslant 0, y \geqslant 0$

Solve: $~$ Associated equation for given constraints are

$x+2 y=10 \text {\ldots (i) }$

$x+y=6 \text { \ldots (i) }$

$3 x+y=8 \text {\ldots (iii) }$

from (i) $\quad x+2 y=10$

$\text { put }$

$y =0 \Rightarrow x=10$

$x =0 \Rightarrow y=5$

$\therefore$ points and (10,0) & (0,5)

from (ii) $\quad x+y=6$

put $\quad x=0 \Rightarrow y=6$

$y=0 \Rightarrow x=6$

$\therefore$ points are (6,0) & (0,6)

from (iii)

$3 x+y=8$

$\text { put } y=0 \Rightarrow x=8 / 3$

Put $y=2, x=2$

put $x=0 \Rightarrow y=8$

$\therefore$ points are (8 / 3,0) & (0,8),(2,2)

$\therefore$ Corner points of the feasible region are

A(10,0), B(2,4),

C(1,5) & D(0,8)

$Z_A=3 \times 10+5 \times 0=30$

$Z_B=3 \times 2+5 \times 4=26 \text { (smallest) }$

$Z_C=3 \times 1+5 \times 5=28$

$Z_D=3 \times 0+5 \times 8=40 \text { (Largest) }$

$3 x+5 y<26$

$3 x+5 y=26$ (Draw this line)

line $\quad 3 x+5 y=26$

have no common points with feasible region

$\therefore \quad Z_{\text {min }}=26 \text { at } B(2,4)$