### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Linear programming <br> lecture-3 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> </div> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ $\rightarrow$ <span style = "color:blue"> Linear programming lecture-3 </span> $\rightarrow$ Corner point method steps $\rightarrow$ Corner point method steps </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Corner point method steps </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - 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 </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ Linear programming lecture-3 $\rightarrow$ <span style = "color:blue"> Corner point method steps </span> $\rightarrow$ Corner point method steps $\rightarrow$ Objective function </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Corner point method steps </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - (d) Obtain value of $z$ at Corner points. - (e) Find $ Z_{max} or Z_{min} $. - (i) Unique - (ii) Line segment- - (iii) No $ Z_{max} / Z_{min}$ (For open region only) </div> </main> <hr> <footer style = "font-size: 18px">Linear programming lecture-3 $\rightarrow$ Corner point method steps $\rightarrow$ <span style = "color:blue"> Corner point method steps </span> $\rightarrow$ Objective function $\rightarrow$ Constraints </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Objective function </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - 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. </div> </main> <hr> <footer style = "font-size: 18px">Corner point method steps $\rightarrow$ Corner point method steps $\rightarrow$ <span style = "color:blue"> Objective function </span> $\rightarrow$ Constraints $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Constraints </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - The inequations or equations on the variables of an $L P P$ are called constrains. They may be of $=,>, \geqslant,<$ or $\leqslant$ type. - The values of the variables $x_1, x_2, \ldots, x_n$ in an LP P are always non-negative. </div> <div style='float: right; width:1%;'> <!--    --> </main> <hr> <footer style = "font-size: 18px">Corner point method steps $\rightarrow$ Objective function $\rightarrow$ <span style = "color:blue"> Constraints </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Problem-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - One kind of cake requires $300 \mathrm{g}$ of flour and $15 \mathrm{g}$ of fat, another kind of cake requires $150 \mathrm{g}$ of flour and $30 \mathrm{g}$ of fat. Find the maximum number of cakes which can be made from $7.5 \mathrm{kg}$ of flour and $600 \mathrm{g}$ of fat. Make it an LPP and solve it graphically. </div> </main> <hr> <footer style = "font-size: 18px">Objective function $\rightarrow$ Constraints $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:20px;'> - Let x and y be no. of cakes of $ {I}^{st}$ and ${II}^{nd}$ kind respectively. | Type | No. of cake | flour | fat | | -------- | -------- | -------- | -------- | | I | x | 300g | 15g | | II | y | 150g | 30g | | | z=x+y | 300x+150y$\leqslant$7500 | 15x+30y$\leqslant$600g| - x$\geqslant$0 , y$\geqslant$0 - let x & y be the no. of ${I}^{st}$ \& ${II}^{nd}$ kind cake respectively. </div> </main> <hr> <footer style = "font-size: 18px">Constraints $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - Formulation as LPP - Maximize $Z=x+y$ (objective function ) - Subject to - $ 300 x+150 y \leqslant 7500 \text { ie. } 2 x+y \leqslant 50$ - $ 15 x+30 y \leqslant 600 \text { ie. } x+2 y \leqslant 40$ - $ x \geqslant 0, y \geqslant 0$ </div> <div style='float: right; width:1%;'> <!--   --> </main> <hr> <footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - 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 $\frac{x}{25}+\frac{y}{50}=1$ </div> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:70%;font-size:30px;'> - 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) }$ </div> <div style='float: right; width:30%;'>  <!----> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:75%;font-size:30px;'> - $ 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 </div> <div style='float: right; width:25%;'>  </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Problem-2 </primary> <hr> <main> <div style='float: left; width:75%;font-size:30px;'> - A furniture deals in only two items table and chair. He has Rs. 10000 to invest and a space to store almost 60 pcs.A table costs him Rs. 500 and a chair costs Rs. 100. He cam sell a table at Rs. 550 and a chair at Rs. 115. Assume that he can sell all the items that he buys. - Formulate this problem as an LPP so that he maximizes his profit. - Solve the problem using corner point method. </div> <div style='float: right; width:1%;'> <!--  --> </div> <!--<div style='float: right; width:20%;'>--> <!----> <!--</div>--> <!----> <!----> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:20px;'> - Let no. of tables = x - no. of chairs = y | Items | no. | Cost | Profit | | -------- | -------- | -------- | -------- | | tables | x | 500 | 50 | | chairs | y | 100 | 15 | | | x=y$\leqslant$ 60 | 500x+100y$\leqslant$10000 | z=50x+15y | - x$\geqslant$0, y$\geqslant$0 - Let $x=$ no. of tables - $ y=$ no. of chairs $\quad$ Formulation as an LPP - Maximize $z=50 x+15 y$ (profit function) </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - 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) </div> <div style='float: right; width:1%;'> <!--   --> </main> <hr> <footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:75%;font-size:30px;'> - 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) </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:60%;font-size:30px;'> - $\therefore$ 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) $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:60%;font-size:30px;'> - $ 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 </div> <div style='float: right; width:20%; max-width:200px;'>  <!----> <!----> </div> <div style='float: right; width:20%;'>  <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Problem-3 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - A manufacturer produces two types of steel trunk. He has two machine $A$ and $B$. The first type of trunk requires $3 \mathrm{hrs}$. on machine $A$ and $3 \mathrm{hrs}$. on machine $B$. The second type of trunk requires $3 \mathrm{hrs}$. on machine $A$ and $2\mathrm{hrs}$. on machine $B$. Machine $A$ an $B$ can work atmost for $18 \mathrm{hrs}$ and $15 \mathrm{hrs}$ per day respectively. He earns a profit of Rs. 30 and Rs. 25 per trunk of the first type and second type respectively. How much trunks of each type must he make each day to make maximum profit? </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:20px;'> - Let $x$ and $y$ be the no. of trunk of first and second type respectively. | Type of trunk | no. of trunk | Machine(A) | Machine(B) | Profit | | -------- | -------- | -------- | -------- | -------- | | I | x | 3 | 3 | 30 | | II | y | 3 | 2 | 25 | | | | 3x+3y$\leqslant$18 | 3x+2y$\leqslant$15 | | - Total Profit, $z=30 x+25 y$ (Profit function) Maximize - subject to constraints </div> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;'> - 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) - <ins>for(i)</ins> $\quad 0+0=0 \leqslant 6$ (True). </div> </main> <hr> <footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:60%;font-size:30px;'> - $\therefore \quad$ 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) - <ins>3</ins> $\quad z=30 x+25 y$ - 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 </div> <div style='float: right; width:0%;'> <!--  --> </div> <div style='float: right; width:25%;'>  <!----> </div> <div style='float: right; width:0%;'> <!--  --> </div> <div style='float: right; width:0%;'>  <!----> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> - 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. </div> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$ </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-3 </Success> ### <primary style="font-size:24px"> Thank you </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;'> </div> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Thank you </span> $\rightarrow$ $\rightarrow$ </footer>