NCERT Solutions For Class 12 Maths Chapter 12 Exercise 12.2
Ex.12.1
Chapter 12 Ex.12.2 Question 1
Reshma wishes to mix two types of food \(P\) and \(Q\) in such a way that the vitamin contents of the mixture contain at least \(8\) units of vitamin \(A\) and \(11\) units of vitamin \(B\). Food \(P\) costs ₹ \(60\)/kg and Food Q costs ₹ \(80\)/kg. Food \(P\) contains \(3\) units/kg of vitamin \(A\) and \(5\) units/kg of vitamin \(B\) while food \(Q\) contains \(4 \)units/kg of vitamin \(A\) and \(2\) units/kg of vitamin \(B\). Determine the minimum cost of the mixture.
Solution
Let the mixture contain \(x\) kg of food \(P\) and \(y\) kg of food \(Q\).
Therefore, \(x \ge 0\) and \(y \ge 0\).
The given information can be shown in a table as follows:
Vitamin A (units/kg) 
Vitamin B (units/kg) 
Cost (₹/kg) 

Food P 
\(3\) 
\(5\) 
\(60\) 
Food Q 
\(4\) 
\(2\) 
\(80\) 
Requirement (units/kg) 
\(8\) 
\(11\) 
The mixture must contain at least \(8\) units of Vitamin \(A\) and \(11\) units of Vitamin \(B\).
Thus, the constraints are
\[\begin{align}3x + 4y &\ge 8\\5x + 2y &\ge 11\end{align}\]
Total cost, \(Z\), of purchasing food is
\[Z = 60x + 80y\]
The mathematical formulation of the given problem is
Minimize
\[Z = 60x + 80y\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&3x + 4y \ge 8 \qquad \qquad \ldots \left( 2 \right)\\&5x + 2y \ge 11\qquad \quad \; \;\ldots \left( 3 \right)\\&x,y \ge 0\qquad \qquad\quad \;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
Since, the feasible region is unbounded.
The corner points of the feasible region are \(A\left( {\frac{8}{3},0} \right),B\left( {2,\frac{1}{2}} \right)\) and \(C\left( {0,\frac{{11}}{2}} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 60x + 80y\) 

\({\rm{A}}\left( {\frac{8}{3},0} \right)\) 
\(160\) 
\(\left. {} \right\} \to \)Minimum 
\({\rm{B}}\left( {2,\frac{1}{2}} \right)\) 
\(160\) 

\({\rm{C}}\left( {0,\frac{{11}}{2}} \right)\) 
\(440\) 
As the feasible region is unbounded, therefore, \(160\) may or may not be the minimum value of \(Z\).
For this, we graph the inequality, \(60x + 80y < 160\) or \(3x + 4y < 8\) and check whether the resulting half plane has points in common with the feasible region or not.
Since, the feasible region has no common point with \(3x + 4y < 8\)
Thus, the minimum cost of the mixture will be ₹ \(160\) at the line segment joining the points \(\left( {\frac{8}{3},0} \right)\) and \(\left( {2,\frac{1}{2}} \right)\).
Chapter 12 Ex.12.2 Question 2
One kind of cake requires \(200\) g flour and \(25\) g of fat, and another kind of cake requires \(100\) g of flour and \(50\) g of fat. Find the maximum number of cakes which can be made from \(5\) kg of flour and \(1\) kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Solution
Let there be \(x\) cakes of first kind and \(y\) cakes of second kind.
Therefore, \(x \ge 0\) and \(y \ge 0\).
The given information can be shown in a table as follows:
Flour (g) 
Fat (g) 

Cakes of first kind, x 
200 
25 
Cakes of second kind, y 
100 
50 
Availability 
5000 
1000 
Hence,
\[\begin{align}&200x + 100y \le 5000\\&\Rightarrow \; 2x + y \le 50\end{align}\]
And
\[\begin{align}&25x + 50y \le 1000\\&\Rightarrow \;x + 2y \le 40\end{align}\]
Total number of cakes, \(Z\), that can be made are,
\[Z = x + y\]
The mathematical formulation of the given problem is
Maximize,
\[Z = x + y\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&2x + y \le 50 \qquad \dots \left( 2 \right)\\&x + 2y \le 40\qquad \dots \left( 3 \right)\\&x,y \ge 0\qquad \quad \;\; \dots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(A\left( {25,0} \right),B\left( {20,10} \right),O\left( {0,0} \right)\) and \(C\left( {0,20} \right)\).
The value of \(Z\) at these corner points are as follows:
Corner point 
\(Z = x + y\) 

\({\rm{A}}\left( {25,0} \right)\) 
25 

\({\rm{B}}\left( {20,10} \right)\) 
30 
\( \to \)Maximum 
\({\rm{O}}\left( {0,0} \right)\) 
0 

\({\rm{C}}\left( {0,20} \right)\) 
20 
Thus, the maximum number of cakes that can be made are \(30\) (\(20\) of one kind and \(10\) of the other kind).
Chapter 12 Ex.12.2 Question 3
A factory makes tennis rackets and cricket bats. A tennis racket takes \(1.5\) hours of machine time and \(3\) hours of craftsman’s time in its making while a cricket bat takes \(3\) hour of machine time and \(1\) hour of craftsman’s time. In a day, the factory has the availability of not more than \(42\) hours of machine time and \(24\) hours of craftsman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is ₹\(20\) and ₹\(10\) respectively, find the maximum profit of the factory when it works at full capacity.
Solution
(i) Let the number of rackets and the number of bats to be made are \(x\) and \(y\), respectively.
The machine time is not available for more than \(42\) hours.
Therefore,
\[1.5x + 3y \le 42\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
The craftsman’s time is not available for more than \(24\) hours.
\[3x + y \le 24\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\]
The factory is to work at full capacity.
Therefore,
\[\begin{align}1.5x + 3y = 42\\3x + y = 24\end{align}\]
On solving these equations, we obtain
\(x = 4\) and \(y = 12\)
Thus, \(4\) rackets and \(12\) bats must be made.
(ii) The given information can be shown in a table as follows:
Tennis Racket 
Cricket Bat 
Availability 

Machine Time (h) 
\(1.5\) 
\(3\) 
\(42\) 
Craftsman’s Time (h) 
\(3\) 
\(1\) 
\(24\) 
Hence,
\[\begin{align}&1.5x + 3y \le 42\\&3x + y \le 24\\&x,\;y \ge 0\end{align}\]
The profit on a racket is ₹ \(20\) and on bat is ₹ \(10\)
Maximize
\[Z = 20x + 10y\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&1.5x + 3y \le 42\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&3x + y \le 24\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\&x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(A\left( {8,0} \right),B\left( {4,12} \right),C\left( {0,14} \right)\) and \(O\left( {0,0} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 20x + 10y\) 

\({\rm{A}}\left( {8,0} \right)\) 
\(160\) 

\({\rm{B}}\left( {4,12} \right)\) 
\(200\) 
\( \to \)Maximum 
\({\rm{C}}\left( {0,14} \right)\) 
\(140\) 

\({\rm{O}}\left( {0,0} \right)\) 
\(0\) 
Thus, the maximum profit of the factory when it works to its full capacity is ₹ \(200.\)
Chapter 12 Ex.12.2 Question 4
A manufacturer produces nuts and bolts. It takes \(1\) hour of work on machine \(A\) and \(3\) hours on machine \(B\) to produce a package of nuts. It takes \(3\) hours on machine \(A\) and \(1\) hour on machine \(B\) to produce a package of bolts. He earns a profit, of ₹ \(17.50\) per package on nuts and ₹ \(7.00\) per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at the most \(12\) hours a day?
Solution
Let the manufacturer produce \(x\) packages of nuts and \(y\) packages of bolts.
Therefore, \(x \ge 0\) and \(y \ge 0\)
The given information can be shown in a table as follows:
Nuts 
Bolts 
Availability 

Machine A (h) 
\(1\) 
\(3\) 
\(12\) 
Machine B (h) 
\(3\) 
\(1\) 
\(12\) 
The profit on a package of nuts is ₹ \(17.50\) and on a package of bolts is ₹ \(7.\)
Therefore, the constraints are
\[\begin{align}x + 3y \le 12\\3x + y \le 12\end{align}\]
Total profit,
\[Z = 17.5x + 7y\]
The mathematical formulation of the given problem is
Maximize
\[Z = 17.5x + 7y\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&x + 3y \le 12\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&3x + y \le 12\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\&x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(O\left( {0,0} \right),A\left( {4,0} \right),B\left( {3,3} \right)\) and \(C\left( {0,4} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 17.5x + 7y\) 

\({\rm{O}}\left( {0,0} \right)\) 
\(0\) 

\({\rm{A}}\left( {4,0} \right)\) 
\(70\) 

\({\rm{B}}\left( {3,3} \right)\) 
\(73.5\) 
\( \to \)Maximum 
\({\rm{C}}\left( {0,4} \right)\) 
\(28\) 
The maximum value of \(Z\) is ₹ \(73.50\) at \(B\left( {3,3} \right)\).
Chapter 12 Ex.12.2 Question 5
A factory manufactures two types of screws, \(A\) and \(B\). Each type of screw requires the use of two machines, an automatic and a hand operated. It takes \(4 \)minutes on the automatic and \(6\) minutes on hand operated machines to manufacture a package of screws A, while it takes \(6\) minutes on automatic and \(3\) minutes on the hand operated machines to manufacture a package of screws \(B\). Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹ \(7\) and screws \(B\) at a prof it of ₹ \(10\). Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Solution
Let the factory manufacture \(x\) screws of type \(A\) and \(y\) screws of type \(B\) on each day.
Therefore, \(x \ge 0\) and \(y \ge 0\)
The given information can be shown in a table as follows:
Screw A 
Screw B 
Availability 

Automatic Machine (min.) 
\(4\) 
\(6\) 
\(4 \times 60 = 240\) 
Hand Operated Machine (min.) 
\(6\) 
\(3\) 
\(4 \times 60 = 240\) 
The profit on a package of screws \(A\) is ₹ \(7\) and on the package of screws \(B\) is ₹ \(10.\)
Therefore, the constraints are
\[\begin{align}4x + 6y \le 240\\6x + 3y \le 240\end{align}\]
Total profit,
\[Z = 7x + 10y\]
The mathematical formulation of the given problem is
Maximize
\[Z = 7x + 10y\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}4x + 6y \le 240\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\6x + 3y \le 240\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(A\left( {40,0} \right),B\left( {30,20} \right)\) and \(C\left( {0,40} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 7x + 10y\) 

\({\text{A}}\left( {40,0} \right)\) 
\(280\) 

\({\rm{B}}\left( {30,20} \right)\) 
\(410\) 
\( \to \)Maximum 
\({\rm{C}}\left( {0,40} \right)\) 
\(400\) 
The maximum value of \(Z\) is \(410\) at \(B\left( {30,20} \right)\).
Chapter 12 Ex.12.2 Question 6
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding /cutting machine and a sprayer. It takes \(2\) hours on grinding /cutting machine and \(3\) hours on the sprayer to manufacture a pedestal lamp. It takes \(1\) hour on the grinding /cutting machine and \(2\) hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most \(20\) hours and the grinding/cutting machine for at the most \(12\) hours. The profit from the sale of a lamp is ₹ \(5\) and that from a shade is ₹ \(3\). Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?
Solution
Let the cottage industry manufacture \(x\) pedestal lamps and \(y\) wooden shades.
Therefore, \(x \ge 0\) and \(y \ge 0\)
The given information can be shown in a table as follows:
Lamps 
Shades 
Availability 

Grinding/Cutting Machine (h) 
\(2\) 
\(1\) 
\(12\) 
Sprayer (h) 
\(3\) 
\(2\) 
\(20\) 
The profit on a lamp is ₹ \(5\) and on the shades is ₹ \(3.\)
Therefore, the constraints are
\[\begin{align}2x + y \le 12\\3x + 2y \le 20\end{align}\]
Total profit,
\[Z = 5x + 3y\]
The mathematical formulation of the given problem is
Maximize
\[Z = 5x + 3y\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&2x + y \le 12\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&3x + 2y \le 20\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\&x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(A\left( {6,0} \right),B\left( {4,4} \right)\) and \(C\left( {0,10} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 5x + 3y\) 

\({\rm{A}}\left( {6,0} \right)\) 
\(30\) 

\({\rm{B}}\left( {4,4} \right)\) 
\(32\) 
\( \to {\rm{Maximum}}\) 
\({\rm{C}}\left( {0,10} \right)\) 
\(30\) 
The maximum value of \(Z\) is \(32\) at \(B\left( {4,4} \right)\).
Therefore, the manufacturer should produce \(4\) pedestal lamps and \(4\) wooden shades to maximize his profits.
Chapter 12 Ex.12.2 Question 7
A company manufactures two types of novelty Souvenirs made of plywood. Souvenirs of type \(A\) require \(5\) minutes each for cutting and \(10\) minutes each for assembling. Souvenirs of type \(B\) require \(8\) minutes each for cutting and \(8\) minutes each for assembling. There are \(3\) hours \(20\) minutes available for cutting and \(4\) hours of assembling. The profit is ₹ \(5\) each for type \(A\) and ₹ \(6\) each for type \(B\) Souvenirs. How many Souvenirs of each type should the company manufacture in order to maximize the profit?
Solution
Let the company manufacture \(x\) Souvenirs of type \(A\) and \(y\) Souvenirs of type \(B.\)
Therefore, \(x \ge 0\) and \(y \ge 0\).
The given information can be shown in a table as follows:
Type A 
Type B 
Availability 

Cutting (min.) 
\(5\) 
\(8\) 
\(3 \times 60 + 20 = 200\) 
Assembling (min.) 
\(10\) 
\(8\) 
\(4 \times 60 = 240\) 
The profit on type \(A\) Souvenirs is ₹ \(5\) and type \(B\) Souvenirs is ₹ \(6\).
Therefore, the constraints are
\[5x + 8y \le 200\]
And
\[\begin{align}10x + 8y \le 240\\ \Rightarrow 5x + 4y \le 120\end{align}\]
Total profit,
\[Z = 5x + 6y\]
The mathematical formulation of the given problem is
Maximize
\[Z = 5x + 6y\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&5x + 8y \le 200\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&5x + 4y \le 120\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\&x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(A\left( {24,0} \right),B\left( {8,20} \right)\) and \(C\left( {0,25} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 5x + 6y\) 

\({\rm{A}}\left( {24,0} \right)\) 
\(120\) 

\({\rm{B}}\left( {8,20} \right)\) 
\(160\) 
\( \to \)Maximum 
\({\rm{C}}\left( {0,25} \right)\) 
\(150\) 
The maximum value of \(Z\) is \(160\) at \(B\left( {8,20} \right)\).
Hence, \(8\) Souvenirs of type \(A\) and \(20\) Souvenirs of type \(B\) should be produced each day to get the maximum profit of ₹ \(160.\)
Chapter 12 Ex.12.2 Question 8
A merchant plans to sell two types of personal computers − a desktop model and a portable model that will cost ₹ \(25000\) and ₹ \(40000\) respectively. He estimates that the total monthly demand of computers will not exceed \(250\) units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ \(70\) lakhs and if h is profit on the desktop model is ₹ \(4500\) and on portable model is ₹ \(5000\).
Solution
Let the merchant stock \(x\) desktop models and \(y\) portable models.
Therefore, \(x \ge 0\) and \(y \ge 0\)
The cost of a desktop model is ₹ \(25000\) and of a portable model is ₹ \(4000\). However, the merchant can invest a maximum of ₹ \(70\) lakhs.
Therefore,
\[\begin{align}&25000x + 40000y \le 7000000\\&\Rightarrow 5x + 8y \le 1400\end{align}\]
The monthly demand of computers will not exceed \(250\) units.
Therefore,
\[x + y \le 250\]
The profit on a desktop model is ₹ \(4500\) and the profit on a portable model is ₹ \(5000\).
Total profit,
\[Z = 4500x + 5000y\]
Thus, the mathematical formulation of the given problem is
Maximize
\[Z = 4500x + 5000y\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&5x + 8y \le 1400\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&x + y \le 250\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\&x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
The corner points are \(A\left( {250,0} \right),B\left( {200,50} \right)\) and \(C\left( {0,175} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 4500x + 5000y\) 

\({\rm{A}}\left( {250,0} \right)\) 
\(1125000\) 

\({\rm{B}}\left( {200,50} \right)\) 
\(1150000\) 
\( \to \)Maximum 
\[{\rm{C}}\left( {0,175} \right)\] 
\(875000\) 
The maximum value of \(Z\) is \(1150000\) at \(B\left( {200,50} \right)\).
Therefore, the merchant should stock \(200\) desktop models and \(50\) portable models to get the maximum profit of ₹ \(1150000.\)
Chapter 12 Ex.12.2 Question 9
A diet is to contain at least \(80\) units of vitamin \(A\) and \(100\) units of minerals. Two foods \({F_1}\) and \({F_2}\) are available. Food \({F_1}\) costs ₹ \(4\) per unit food and \({F_2}\) costs ₹ \(6\) per unit. One unit of food \({F_1}\) contains \(3\) units of vitamin \(A\) and \(4\) units of minerals. One unit of food \({F_2}\) contains \(6\) units of vitamin \(A\) and \(3\) units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Solution
Let the diet contain \(x\) units of food \({F_1}\) and y units of food \(F_2\).
Therefore,\(x \ge 0\) and \(y \ge 0\)
The given information can be shown in a table as follows:
Vitamin A (units) 
Mineral (units) 
Cost per unit (?) 

Food F_{1} (x) 
\(3\) 
\(4\) 
\(4\) 
Food F_{2} (y) 
\(6\) 
\(3\) 
\(6\) 
Requirement 
\(80\) 
\(100\) 
The cost of food \({F_1}\) is ₹ \(4/\rm{unit}\) and of food \({F_2}\) is ₹ \(6/\rm{unit.}\)
Therefore, the constraints are
\[\begin{align}3x + 6y \ge 80\\4x + 3y \ge 100\\x,\;y \ge 0\end{align}\]
Total cost of the diet,
\[Z = 4x + 6y\]
The mathematical formulation of the given problem is
Minimize
\[Z = 4x + 6y\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}3x + 6y \ge 80\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\4x + 3y \ge 100\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the constraints is given by
Since, the feasible region is unbounded.
The corner points of the feasible region are \(A\left( {\frac{{80}}{3},0} \right),B\left( {24,\frac{4}{3}} \right)\) and \(C\left( {0,\frac{{100}}{3}} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 4x + 6y\) 

\({\rm{A}}\left( {\frac{{80}}{3},0} \right)\) 
\(\frac{{320}}{3} = 106.67\) 

\({\rm{B}}\left( {24,\frac{4}{3}} \right)\) 
\(104\) 
\( \to \)Minimum 
\({\rm{C}}\left( {0,\frac{{100}}{3}} \right)\) 
\(200\) 
As the feasible region is unbounded, therefore, \(104\) may or may not be the minimum value of \(Z.\)
For this, we draw a graph of the inequality, \(4x + 6y < 104\) or \(2x + 3y < 52\), and check whether the resulting half plane has points in common with the feasible region or not.
Since, the feasible region has no common point with \(2x + 3y < 52\).
Therefore, the minimum cost of the mixture will be ₹ 104.
Chapter 12 Ex.12.2 Question 10
There are two types of fertilizers \({F_1}\) and \({F_2}\). \({F_1}\) consists of 10% nitrogen and 6% phosphoric acid and \({F_2}\) consists of \(5\%\) nitrogen and \(10\%\) phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least \(14\rm{ kg}\) of nitrogen and \(14\rm{ kg}\) of phosphoric acid for her crop. If \({F_1}\) cost ₹ \(6/\rm{kg}\) and \({F_2}\) costs ₹ \(5/\rm{kg}\), determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution
Let the farmer buy \(x\,\rm{kg}\) of fertilizer \({F_1}\) and \(y\,\rm{kg}\) of fertilizer\({F_2}\).
Therefore, \(x \ge 0\) and \(y \ge 0\)
The given information can be shown in a table as follows:
Nitrogen (%) 
Phosphoric Acid (%) 
Cost (?/kg) 

F_{1} (x) 
\(10\) 
\(6\) 
\(6\) 
F_{2} (y) 
\(5\) 
\(10\) 
\(5\) 
Requirement (kg) 
\(14\) 
\(14\) 
\({F_1}\) consists of \(10\% \)nitrogen and \({F_2}\) consists of \(5\%\) nitrogen. However, the farmer requires at least \(14\,\rm {kg}\) of nitrogen.
Therefore,
\[\begin{align}&10\% \;of\;x + 5\% \;of\;y \ge 14\\&\frac{x}{{10}} + \frac{y}{{20}} \ge 14\\&2x + y \ge 280\end{align}\]
\({F_1}\) consists of \(6\%\) phosphoric acid and \({F_2}\) consists of \(10\%\) phosphoric acid.
However, the farmer requires at least \(14\rm{ kg}\) of phosphoric acid.
Therefore,
\[\begin{align}&6\% \;of\;x + 10\% \;of\;y &\ge 14\\&\frac{{6x}}{{100}} + \frac{{10y}}{{100}} &\ge 14\\&3x + 5y &\ge 700\end{align}\]
Total cost of fertilizers,
\[Z = 6x + 5y\]
The mathematical formulation of the given problem is
Minimize
\[Z = 6x + 5y\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]
Subject to the constraints,
\[\begin{align}&2x + y \ge 280\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&3x + 5y \ge 700\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\&x,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]
The feasible region determined by the system of constraints is given by
Since, the feasible region is unbounded.
The corner points are \(A\left( {\frac{{700}}{3},0} \right),B\left( {100,80} \right)\) and \(C\left( {0,280} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 6x + 5y\) 

\({\rm{A}}\left( {\frac{{700}}{3},0} \right)\) 
\(1400\) 

\({\rm{B}}\left( {100,80} \right)\) 
\(1000\) 
\( \to \)Minimum 
\({\rm{C}}\left( {0,280} \right)\) 
\(1400\) 
As the feasible region is unbounded, therefore, \(1000\) may or may not be the minimum value of \(Z.\)
For this, we draw a graph of the inequality, \(6x + 5y < 1000\), and check whether the resulting half plane has points in common with the feasible region or not.
Since, the feasible region has no common point with \(6x + 5y < 1000\).
Thus, \(100\,\rm{ kg}\) of fertilizer \({F_1}\) and \(80 \,\rm{kg}\) of fertilizer \({F_2}\) should be used to minimize the cost. The minimum cost is ₹ \(1000.\)
Chapter 12 Ex.12.2 Question 11
The corner points of the feasible region determined by the following system of linear inequalities:
\(2x + y \le 10,\;x + 3y \le 15,\;x,y \ge 0\) are \(\left( {0,0} \right),\;\left( {5,0} \right),\;\left( {3,4} \right)\) and \(\left( {0,5} \right)\). Let \(Z = px + qy,\) where\(p,\;q > 0\). Condition on \(p\) and \(q\) so that the maximum of \(Z\) occurs at both\(\left( {3,4} \right)\) and \(\left( {0,5} \right)\) is
(A) \(p = q\)
(B) \(p = 2q\)
(C) \(p = 3q\)
(D) \(q = 3p\)
Solution
The maximum value of \(Z\) is unique.
It is given that the maximum value of \(Z\) occurs at two points, \(\left( {3,4} \right)\) and \(\left( {0,5} \right)\).
Therefore, Value of Z at \(\left( {3,4} \right) = \) Value of Z at \(\left( {0,5} \right)\)
\[\begin{align}&\Rightarrow\; p\left( 3 \right) + q\left( 4 \right) = p\left( 0 \right) + q\left( 5 \right)\\&\Rightarrow \;3p + 4q = 5q\\&\Rightarrow\; q = 3p\end{align}\]
Thus, the correct option is (D).