\displaystyle \textbf{Question 1: } \text{Maximise } Z=4x-3y,   \text{ Subject to the following constraints}
\displaystyle 2x+y\leq20,\;3x-2y\geq16,\;x\leq9,\;y\geq1
\displaystyle \text{Also, find } Z\text{'s maximum value.}
\displaystyle \text{Answer:}
\displaystyle  \text{We have following LPP}
\displaystyle \text{Maximise } Z=4x-3y
\displaystyle \text{Subjected to the constraints}
\displaystyle 2x+y\leq20,\ 3x-2y\geq16,
\displaystyle x\leq9\text{ and }y\geq1
\displaystyle \text{Now, considering the equations as equations, we get}
\displaystyle 2x+y=20 \qquad \text{(i)}
\displaystyle 3x-2y=16 \qquad \text{(ii)}
\displaystyle \text{Table for line }2x+y=20\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 10 \\ \hline y & 20 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line is passing through the points }(0,20)\text{ and }(10,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+y\leq20,\text{ we get}
\displaystyle 0+0\leq20,\text{ which is true.}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Table for line }3x-2y=16\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & \frac{16}{3} \\ \hline y & -8 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line is passing through the points }(0,-8)\text{ and }\left(\frac{16}{3},0\right).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x-2y\geq16,\text{ we get}
\displaystyle 0-0\geq16,\text{ which is false.}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{The graphical representation of the system of inequalities is as given below}
\displaystyle \text{Clearly, the feasible region is }ABCDA,\text{ where the corner points are }A(6,1),\ B(9,1),\ C(9,2)\text{ and }D(8,4).
\displaystyle \text{Now, the values of }Z\text{ at corner points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner\ points} & Z=4x-3y \\ \hline A(6,1) & Z=24-3=21 \\ \hline B(9,1) & Z=36-3=33\ (\text{Maximum}) \\ \hline C(9,2) & Z=36-6=30 \\ \hline D(8,4) & Z=32-12=20 \\ \hline \end{array}
\displaystyle \text{Hence, the maximum value of }Z\text{ is }33,\text{ which occurs at }(9,1).
\\

\displaystyle \textbf{Question 2: } \text{A soap company manufactures two varieties of soap using two machines. The time taken}
\displaystyle \text{on each machine to manufacture one batch of each variety of soap is given below.}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Soap Variety} & \text{Machine 1} & \text{Machine 2} \\ \hline \text{Soap A} & 1\text{ h} & 1\text{ h} \\ \hline \text{Soap B} & 2\text{ h} & 1\text{ h} \\ \hline \end{array}
\displaystyle \text{Machine 1 and Machine 2 cannot be operational for more than }12\text{ hours and }10\text{ hours in a}
\displaystyle \text{day, respectively. The profit earned from one batch of Soap A and one batch of Soap B is}
\displaystyle \text{Rs }3000\text{ and Rs }4000\text{ respectively.}
\displaystyle \text{Find the number of batches of each kind of soap that should be made daily to maximise the}
\displaystyle \text{profit.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the number of batches of Soap }A\text{ be }x\text{ and of Soap }B\text{ be }y.
\displaystyle \therefore \text{Objective function is maximise,}
\displaystyle Z=3000x+4000y
\displaystyle \text{Machine 1 cannot be operational for more than }12\text{ h.}
\displaystyle \text{So, first constraint is }x+2y\leq12.
\displaystyle \text{Machine 2 cannot be operational for more than }10\text{ h.}
\displaystyle \text{So, second constraint is }x+y\leq10.
\displaystyle \text{Also, the number of batches of each kind of soap cannot be negative.}
\displaystyle \text{So, non-negative restrictions or constraints are}
\displaystyle x\geq0,\ y\geq0
\displaystyle \text{Hence, the linear programming problem for the given problem is as follows :}
\displaystyle \text{Maximise } Z=3000x+4000y
\displaystyle \text{Subjected to constraints}
\displaystyle x+2y\leq12,
\displaystyle x+y\leq10,
\displaystyle x\geq0\text{ and }y\geq0
\displaystyle \text{Now, table for the line }x+2y=12\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 12 \\ \hline y & 6 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line is passing through the points }(0,6)\text{ and }(12,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\leq12,\text{ we get}
\displaystyle 0+0\leq12,\text{ which is true.}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Table for the line }x+y=10\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 10 \\ \hline y & 10 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line is passing through the points }(0,10)\text{ and }(10,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\leq10,\text{ we get}
\displaystyle 0+0\leq10,\text{ which is true.}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Since, }x\geq0\text{ and }y\geq0
\displaystyle \text{So, the feasible region lies in the first quadrant.}
\displaystyle \text{On solving equations }x+2y=12\text{ and }x+y=10,\text{ we get}
\displaystyle B(8,2).
\displaystyle \text{The graphical representation of the obtained system of inequalities is as follows.}
\displaystyle \text{Clearly, the feasible region is }OABCO.
\displaystyle \text{The corner points of the feasible region are }O(0,0),
\displaystyle A(10,0),\ B(8,2)\text{ and }C(0,6).
\displaystyle \text{Now, the value of }Z\text{ at these points are as follows.}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner\ points} & Z=3000x+4000y \\ \hline O(0,0) & Z=0+0=0 \\ \hline A(10,0) & Z=30000+0=30000 \\ \hline B(8,2) & Z=24000+8000=32000\ (\text{maximum}) \\ \hline C(0,6) & Z=0+24000=24000 \\ \hline \end{array}
\displaystyle \text{Thus, the maximum value of }Z\text{ is Rs }32000,\text{ which occurs at }B(8,2).
\displaystyle \text{Hence, }8\text{ batches of Soap }A\text{ and }2\text{ batches of Soap }B
\displaystyle \text{should be made daily to maximise the profit.}
\\

\displaystyle \textbf{Question 3: } \text{The standard weight of a special purpose brick is }5\text{ kg and it must contain}
\displaystyle \text{two basic ingredients }B_1\text{ and }B_2.\;B_1\text{ costs Rs }5\text{ per kg and }B_2\text{ costs Rs }8\text{ per kg.}
\displaystyle \text{Strength considerations dictate that the brick should not contain more than }4\text{ kg of }B_1
\displaystyle \text{and minimum }2\text{ kg of }B_2.\text{ Since, the demand for the product is likely to be related to the}
\displaystyle \text{price of the bricks, find the minimum cost of the brick satisfying the above conditions.}
\displaystyle \text{Formulate this situation as an LPP and solve it graphically.} 
\displaystyle \text{Answer:}
\displaystyle  \text{Let }x\text{ kg of }B_1\text{ and }y\text{ kg of }B_2\text{ are taken for making brick.}
\displaystyle \text{Here, } Z=5x+8y\text{ is the cost which is objective}
\displaystyle \text{function and is to be maximise subject to constraints}
\displaystyle x+y=5 \qquad \text{(i)}
\displaystyle x\leq4 \qquad \text{(ii)}
\displaystyle y\geq2 \qquad \text{(iii)}
\displaystyle x,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{In this case constraints (i) is a line passing through the}
\displaystyle \text{points }(5,0)\text{ and }(0,5).
\displaystyle \text{Therefore, maximum or minimum value of objective}
\displaystyle \text{function }'Z'\text{ exist on end points of line (constraints) (i)}
\displaystyle \text{in feasible region i.e. at }A\text{ or }B.
\displaystyle \text{At }A(3,2),\ Z=5\times3+8\times2=31\ (\text{minimum})
\displaystyle \text{At }B(0,5),\ Z=5\times0+8\times5=40\ (\text{maximum})
\displaystyle \text{Hence, cost of brick is minimum when }3\text{ kg of }B_1\text{ and}
\displaystyle 2\text{ kg of }B_2\text{ are taken.}
\\

\displaystyle \textbf{Question 4: } \text{A manufacturer has two machines }X\text{ and }Y\text{ that may run at the most}
\displaystyle 360 \text{min in a day to produce two types of toys }A\text{ and }B.
\displaystyle \text{To produce each toy }A,\text{ machines }X\text{ and }Y\text{ need to run at the most }12\text{ min and }6\text{ min,}
\displaystyle \text{respectively. To produce each toy }B,\text{ machines }X\text{ and }Y\text{ need to run at the most }6\text{ min}
\displaystyle \text{and }9\text{ min, respectively.}
\displaystyle \text{By selling the toys }A\text{ and }B,\text{ the manufacturer makes the profits of Rs }30\text{ and Rs }20,
\displaystyle \text{respectively.}
\displaystyle \text{Formulate a linear programming problem and find the number of toys }A\text{ and }B\text{ that should}
\displaystyle \text{be manufactured in a day to get maximum profit.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }x\text{ and }y\text{ be the number of toys of type }A\text{ and}
\displaystyle \text{type }B\text{ respectively.}
\displaystyle \text{The data given in the problem can be summarized in}
\displaystyle \text{the following tabular form.}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Type of toys} & \text{Machine X} & \text{Machine Y} \\ \hline A & 12 & 6 \\ \hline B & 6 & 9 \\ \hline \text{Maximum time available} & 360 & 360 \\ \hline \end{array}
\displaystyle \text{Since, the profit is to be maximized.}
\displaystyle \therefore \text{Maximise, } Z=30x+20y
\displaystyle \text{Subject to the constraints are}
\displaystyle 12x+6y\leq360 \qquad \text{(i)}
\displaystyle 6x+9y\leq360 \qquad \text{(ii)}
\displaystyle x\geq0, \qquad \text{(iii)}
\displaystyle y\geq0 \qquad \text{(iv)}
\displaystyle \text{Table for graph }12x+6y\leq360
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 30 \\ \hline y & 60 & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }12x+6y\leq360,\text{ we get}
\displaystyle 0+0\leq360 \Rightarrow 0\leq360,\text{ which is true.}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for graph }6x+9y\leq360
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 60 \\ \hline y & 40 & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the equality }6x+9y\leq360,\text{ we get}
\displaystyle 0+0\leq360 \Rightarrow 0\leq360,\text{ which is true.}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0.\text{ So, the feasible region lies in the first}
\displaystyle \text{quadrant.}
\displaystyle \text{On solving Eqs. (i) and (ii), we get intersection point}
\displaystyle C(15,30).
\displaystyle \text{The shaded region represents the feasible region}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z=30x+20y \\ \hline O(0,0) & Z=30\times0+20\times0=0 \\ \hline A(30,0) & Z=30\times30+20\times0=900 \\ \hline B(0,40) & Z=30\times0+20\times40=800 \\ \hline C(15,30) & Z=30\times15+20\times30=1050 \\ \hline \end{array}
\displaystyle \text{Maximum profit is Rs }1050\text{ at }C(15,30).
\displaystyle \text{Type }A=15\text{ toys and type }B=30\text{ toys should be}
\displaystyle \text{manufactured in a day to maximize the profit.}
\\

\displaystyle \textbf{Question 5: } \text{A manufacturer wishes to produce two commodities }A\text{ and }B.\text{ The}
\displaystyle \text{number of units of material, labour and equipment needed to produce one unit of each}
\displaystyle \text{commodity is shown in the table given below. Also show the available number of}
\displaystyle \text{units of each item, material, labour, and equipment.}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Items} & \text{Commodity }A & \text{Commodity }B & \text{Available Number of Units} \\ \hline \text{Material} & 1 & 2 & 8 \\ \hline \text{Labour} & 3 & 2 & 12 \\ \hline \text{Equipment} & 1 & 1 & 10 \\ \hline \end{array}
\displaystyle \text{Find the maximum profit, if each unit of commodity }A\text{ earns a profit of Rs }2\text{ and each}
\displaystyle \text{unit of commodity }B\text{ earns a profit of Rs }3.   
\displaystyle \text{Answer:}
\displaystyle  \text{Let }x\text{ be the number of units of commodity }A\text{ produced}
\displaystyle \text{and }y\text{ be the number of units of commodity }B\text{ produced.}
\displaystyle \text{Maximise, } Z=2x+3y
\displaystyle \therefore x+2y\leq8 \qquad [\text{material constraint}]
\displaystyle 3x+2y\leq12 \qquad [\text{labour constraint}]
\displaystyle x+y\leq10 \qquad [\text{equipment constraint}]
\displaystyle x,\ y\geq0 \qquad [\text{non-negative constraint}]
\displaystyle \text{Now, consider the inequalities as equation}
\displaystyle \therefore \text{Table for line }x+2y=8
\displaystyle \begin{array}{|c|c|c|} \hline x & 8 & 0 \\ \hline y & 0 & 4 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in }x+2y\leq8,\text{ we get}
\displaystyle 0+0\leq8,\text{ which is true}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Table for line }3x+2y=12
\displaystyle \begin{array}{|c|c|c|} \hline x & 4 & 0 \\ \hline y & 0 & 6 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in }3x+2y\leq12,\text{ we get}
\displaystyle 0+0\leq12,\text{ which is true}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Table for line }x+y=10
\displaystyle \begin{array}{|c|c|c|} \hline x & 10 & 0 \\ \hline y & 0 & 10 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in }x+y\leq10,\text{ we get}
\displaystyle 0+0\leq10,\text{ which is true}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Now, we have to maximize } Z=2x+3y
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Value of } Z=2x+3y \\ \hline (0,0) & 0 \\ \hline (4,0) & 8 \\ \hline (2,3) & 13\ (\text{Maximum}) \\ \hline (0,4) & 12 \\ \hline \end{array}
\displaystyle \text{Hence, the maximum profit is Rs }13\text{ at }(2,3).
\\

\displaystyle \textbf{Question 6: } \text{A farm is engaged in breeding hens. The hens are fed products }A\text{ and }B
\displaystyle \text{grown in the farm which contains nutrients }P,\;Q\text{ and }R.\text{ One kilogram of}
\displaystyle \text{product }A\text{ contains }36\text{ units, }3\text{ units and }20\text{ units of nutrients }P,\;Q\text{ and }R
\displaystyle \text{respectively, whereas one kilogram of product }B\text{ contains }6\text{ units, }12\text{ units}
\displaystyle \text{and }10\text{ units of nutrients }P,\;Q\text{ and }R\text{ respectively.}
\displaystyle \text{The minimum requirement of nutrients }P,\;Q\text{ and }R\text{ for a hen is }108,\;36\text{ and }100
\displaystyle \text{units respectively. Product }A\text{ costs Rs }20\text{ per kilogram and product }B\text{ costs}
\displaystyle \text{Rs }40\text{ per kilogram. Using linear programming, find the number of kilograms of}
\displaystyle \text{products }A\text{ and }B\text{ to be produced to minimize the total cost. Identify the feasible}
\displaystyle \text{region from the rough sketch.} 
\displaystyle \text{Answer:}
\displaystyle  \text{The data given in the problem can be summarised in the following tabular form.}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Product} & \multicolumn{3}{c|}{\text{Nutrient Constituent}} & \text{Cost in (Rs)} \\ \cline{2-4} & P & Q & R & \\ \hline A & 36 & 3 & 20 & 20 \\ \hline B & 6 & 12 & 10 & 40 \\ \hline \text{Minimum Requirement} & 108 & 36 & 100 & \\ \hline \end{array}
\displaystyle \text{Let }x\text{ units of product }A\text{ and }y\text{ units of product }B\text{ be bought to fulfil}
\displaystyle \text{the minimum requirement of }P,\;Q\text{ and }R\text{ and to minimise the cost.}
\displaystyle \text{The mathematical formulation of the above problem is as follows:}
\displaystyle \text{Minimise } Z=20x+40y
\displaystyle \text{Subject to }36x+6y\geq108
\displaystyle 3x+12y\geq36
\displaystyle 20x+10y\geq100
\displaystyle x\geq0,\;y\geq0
\displaystyle \text{The set of all feasible solutions of the above LPP is represented by the feasible}
\displaystyle \text{region shaded in the figure.}
\displaystyle \text{The coordinates of the corner points of the feasible region are}
\displaystyle A_2(12,0),\;P_1(4,2),\;P_2(2,6)\text{ and }B_1(0,18).
\displaystyle \text{Now, we have to find a point in the feasible region which gives the minimum}
\displaystyle \text{value of the objective function.}
\displaystyle \text{For this, let us give some value to }Z,\text{ say }20,\text{ and draw a dotted line}
\displaystyle 20x+40y=20.
\displaystyle \text{Now, draw lines parallel to this line which have at least one point common}
\displaystyle \text{to the feasible region and locate the line which is nearest to the origin and}
\displaystyle \text{has at least one point common to the feasible region.}
\displaystyle \text{Clearly, such a line is }Z_1=20x+40y\text{ and it has the point }P_1(4,2)\text{ common}
\displaystyle \text{with the feasible region.}
\displaystyle \text{Thus, }Z_1=20x+40y\text{ is the minimum value of }Z,\text{ and the feasible solution}
\displaystyle \text{which gives this value of }Z\text{ is the corner point }P_1(4,2)\text{ of the shaded region.}
\displaystyle \text{The values of the variables for the optimal solution are }x=4,\;y=2.
\displaystyle \text{Substituting these values in }Z=20x+40y,\text{ we get}
\displaystyle Z=160\text{ as the optimal value of }Z.
\displaystyle \text{Hence, }4\text{ units of product }A\text{ and }2\text{ units of product }B\text{ are sufficient}
\displaystyle \text{to fulfil the minimum requirement at a minimum cost of Rs }160.
\\

\displaystyle \textbf{Question 7: } \text{A toy company manufactures two types of dolls }A\text{ and }B.\text{ Market test and}
\displaystyle \text{available resources have indicated that the combined production level should not exceed}
\displaystyle 1200\text{ dolls per week and the demands for the dolls of type }B\text{ is at most half of that}
\displaystyle \text{for dolls of type }A.
\displaystyle \text{Further, the production level of type }A\text{ can exceed three times the production of dolls}
\displaystyle \text{of other type by at most }600\text{ units. If the company makes profit of Rs }12\text{ and Rs }16
\displaystyle \text{per doll respectively on dolls }A\text{ and }B,\text{ how many of each type of dolls should be}
\displaystyle \text{produced weekly, in order to maximise the profit?} 
\displaystyle \text{Answer:}
\displaystyle  \text{Let the company manufactures }x\text{ dolls of type }A\text{ and }y
\displaystyle \text{dolls of type }B,\text{ then}
\displaystyle x\geq0,\ y\geq0 \qquad \text{(i)}
\displaystyle x+y\leq1200 \qquad \text{(ii)}
\displaystyle y\leq\frac{x}{2}\Rightarrow x-2y\geq0 \qquad \text{(iii)}
\displaystyle x\leq3y+600\Rightarrow x-3y\leq600 \qquad \text{(iv)}
\displaystyle \text{Let }Z\text{ be the total profit, then}
\displaystyle Z=12x+16y
\displaystyle \text{Firstly, draw the graph of the line }x+y=1200
\displaystyle \begin{array}{|c|c|c|} \hline x & 1200 & 0 \\ \hline y & 0 & 1200 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\leq1200,\text{ we get}
\displaystyle 0+0\leq1200 \Rightarrow 0\leq1200,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Secondly, draw the graph of line }x-2y=0
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 2 \\ \hline y & 0 & 1 \\ \hline \end{array}
\displaystyle \text{On putting }(200,0)\text{ in the inequality }x-2y\geq0,\text{ we}
\displaystyle \text{have.}
\displaystyle 200-2\times0\geq0 \Rightarrow 200\geq0,\text{ which is true}
\displaystyle \text{So, the half plane is towards the }X\text{-axis.}
\displaystyle \text{Thirdly, draw the graph of the line }x-3y=600
\displaystyle \begin{array}{|c|c|c|} \hline x & 600 & 0 \\ \hline y & 0 & -200 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x-3y\leq600,\text{ we get}
\displaystyle 0-3\times0\leq600 \Rightarrow 0\leq600,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Since, }x,\ y\geq0.
\displaystyle \text{So, the feasible region lies in the first quadrant.}
\displaystyle \text{The point of intersection of lines }x-3y=600\text{ and}
\displaystyle x+y=1200\text{ is }B(1050,150),\text{ lines }x=2y\text{ and}
\displaystyle x+y=1200\text{ is }C(800,400).
\displaystyle \therefore \text{Feasible region is }OABCO.
\displaystyle \text{The corner points of the feasible region are }O(0,0),
\displaystyle A(600,0),\ B(1050,150)\text{ and }C(800,400).
\displaystyle \text{The values of }Z\text{ at these points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner point} & Z=12x+16y \\ \hline O(0,0) & 0 \\ \hline A(600,0) & 7200 \\ \hline B(1050,150) & 15000 \\ \hline C(800,400) & 16000=\text{maximum} \\ \hline \end{array}
\displaystyle \text{The maximum value of }Z\text{ is }16000\text{ at }C(800,400).
\displaystyle \text{Thus, }800\text{ and }400\text{ dolls of type }A\text{ and type }B\text{ should}
\displaystyle \text{be produced respectively, so that company get the}
\displaystyle \text{maximum profit of Rs }16000.
\\

\displaystyle \textbf{Question 8: } \text{Solve the following linear programming problem graphically and interpret}
\displaystyle \text{your result of} Z=2x-5y\text{ subject to the constraints } \\ x+y\geq2,\;x-y\geq0,\;x\leq1,\;x\geq0,\;y\geq0.
\displaystyle \text{Answer:}
\displaystyle  \text{Given, objective function } Z=2x-5y
\displaystyle \text{Subject to constraints,}
\displaystyle x+y\geq2 \qquad \text{(i)}
\displaystyle x-y\geq0 \qquad \text{(ii)}
\displaystyle x\leq1,\ x\geq0,\ y\geq0 \qquad \text{(iii)}\displaystyle \text{and table for line }x+y=2
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 2 \\ \hline y & 2 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line is passing through the points }(2,0)\text{ and }(0,2).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\geq2,\text{ we get}
\displaystyle 0+0\geq2,\text{ which is false.}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for line }x-y\geq0
\displaystyle \begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline y & 0 & 1 & 2 \\ \hline \end{array}
\displaystyle \text{So, the line is passing through the origin.}
\displaystyle \text{For }x\leq1,\text{ the half plane is toward the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0,\text{ implies that the region lies in the}
\displaystyle \text{1st quadrant.}
\displaystyle \text{On drawing the graph there will no common region}
\displaystyle \text{bounded and }C(1,1)\text{ is the only common point in the}
\displaystyle \text{region.}
\displaystyle \text{So, } Z=2\times1-5\times1=2-5=-3.
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.