CORNER POINT METHOD

STEPS:

$\rightarrow$ FORMULATION OF LPP

(a) (i) Objective Function (To be Maximize or Minimize)

$z=a x+b y$

(ii) Linear Constraints

(b) Graphical Representation of Constraints to obtain feasible region (open/closed)

(c) Corner points of feasible region

If $a_1, a_3, a_2, \ldots, a_n$ are constraints and

$x_1, x_2, x_3, \ldots, x_n$ are variables (called decision variables), then linear function

$z=a_1 x_1+a_2 x_2+a_2 x_3+\cdots+a_n x_n$

Which is to be optimized is called the objective function.

It is always non negative.

Constraints are

$2 x+y \leqslant 50 \ldots { (i) }$

$x+2 y \leqslant 40 \ldots { (ii) }$

Associated equation for (i) & (ii) is

$2 x+y=50 \ldots{ (i) }$

$x+2 y=40 \ldots { (ii) }$

for (i) put $\frac{x}{25}+\frac{y}{50}=1$

for (ii) $\quad \frac{x}{40}+\frac{y}{20}=1$

origin test for (i)

$2 \times 0+0=0 \leqslant 50 \text { (True). }$

origin test for (ii)

$0+2 \times 0=0 \leqslant 40 \text { (True) }$

$z=x+y$

$Z_A=25+0=25$

$Z_B=20+10=30$

$Z_C=0+20=20$

$Z$ is maximum at $B(20,10)$

$\therefore \text{no. of } {I}^{st}$ cake = 20

${II}^{nd}$ cake = 10

Subject to constraints

$5 x+y \leqslant 100$ (Investment constraints)

$x+y \leqslant 60$ (storage constraints)

$x \geqslant 0, y \geqslant 0$ (non-negative constraints)

Linear constraints an

$5 x+y \leqslant 100 \ldots { (i) }$ $x+y \leqslant 60 \ldots {(ii) }$

Associated equation for (i) & (ii)

$5 x+y=100 \Rightarrow \frac{x}{20}+\frac{y}{100}=1$

$x+y=60 \Rightarrow \frac{x}{60}+\frac{y}{60}=1$

origin test for (i) 5 $\times 0+0=0\leqslant$ 100(true)

$\therefore$ origin lies in the solution region of (i) origin test for (ii) $0+0\leqslant$60(true)

$\therefore$ origin lies in the solution region $4z=50 x+15 y$

$\text{Corner Points are}$ $A(20,0)$ $B(10,50)$ $C(0,60)$

$Z_A=50 \times 20+15 \times 0$ $=1000$

$Z_B=50 \times 10+15 \times 50$ $=1250$

$Z_c =50 \times 0+15 \times 60 =900$

Maximum value of $z=1250$ at $\quad$ $B(10,50)$

$\therefore$ no. of tables $=10$ $\quad$ no. of chairs = 50

Total Profit, $z=30 x+25 y$ (Profit function) Maximize

subject to constraints

Subject to constraints

$3 x+3 y \leqslant 18 \text { ie. } x+y \leqslant 6 \ldots (i)$

$3 x+2 y \leqslant 15 \ldots (ii)$

$x, y \geqslant 0$

Associated equation for (i) & (ii)

$x+y=6 \Rightarrow \frac{x}{6}+\frac{y}{6}=1$

$3 x+2 y=15 \Rightarrow \frac{x}{5}+\frac{y}{7 \cdot 5}=1$

Origin test for (i) & (ii)

for(i) $\quad 0+0=0 \leqslant 6$ (True).

$\therefore \quad$ origin lies in the solution region of (i)

$\quad$ origin test for (ii)

$3 \times 0+2 \times 0=0 \leqslant 15$( true)

$\therefore$ origin lies in the solution region for (ii)

3 $\quad z=30 x+25 y$

Corner points $A(5,0), ~ B(3,3), ~ C(0,6)$

$Z_A =30 \times 5+25 \times 0$ =150

$Z_B =30 \times 3+25 \times 3$ =165

$Z_C =30 \times 0+25 \times 6$ =150

So, maximum values of profit function

$Z=30 x+25 y \text { occurs at } B(3,3)$

$\therefore Z_{\text {max }}=165 \text { at } B(3,3)$

$\therefore$ Manufacturer should produce 3 trunks of each type to get maximum profit of Rs. 165.