Linear Programming

Minimize $z=3x+5y$

Such that $x+3y\ge 3,x+y\ge 2,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+3y=3\\ x+y=2\\ x,y=0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{3}+\frac{y}{1}=1\\ \frac{x}{2}+\frac{y}{2}=1\\ x,y=0\end{array}$

The graph of the given inequalities is

The feasible region is unbounded. The corner points are $A\left(3,0\right),B\left(\frac{3}{2},\frac{1}{2}\right)\&C\left(0,2\right)$

The values of Z at these corner points as follows.

As the feasible region is unbounded, 7 may or may not be minimum v

Maximise $z=5x+3y$

Subject to $2x+5y\le 15,5x+2y\le 10,x\ge 0,y\ge 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l}3x+5y=15\\ 5x+2y=10\&x=0,y=0\end{array}$

$\Rightarrow \frac{x}{5}+\frac{y}{3}=1$

$\begin{array}{l}\frac{x}{2}+\frac{y}{5}=1\\ x=0,y=0\end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O\left(0,0\right),A\left(2,0\right),B\left(\frac{20}{19},\frac{45}{19}\right)\&C\left(0,3\right)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19}at,B\left(\frac{20}{19},\frac{45}{19}\right)$

Minimize $z=-3x+4y$

Subject to $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=8\\ 3x+2y=12\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{8}+\frac{y}{4}=1$

$\begin{array}{l}\frac{x}{4}+\frac{y}{6}=1\\ x=0,y=0\end{array}$

The graph is shown below.

The bounded region OABC is the feasible region with the corner points O (0,0), A (4,0), B (2,3), and C (0,4

The value of Z at these points are

Therefore, the minimum value of Z is -12 at (4,0

Maximise $z=3x+4y$

Subject to the constraints: $x+y\le 4,x\ge 0,y\ge 0$

The corresponding equation of the above inequality are

$\begin{array}{l}x+y=4\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{4}+\frac{y}{4}=1$

$x=0,y=0$

The graph of the given inequalities.

The shaded region OAB is the feasible region which is bounded.

The corresponding of the corner point of the feasible region are O (0,0), A (4,0), and B (0,4).

The va

The graphical method in Linear Programmin is a visual approach used to find the optimal solution of questions in Linear Programming. We have provided NCERT Solutions for Class 12 Linear Programming and we have discussed graphical solutions of all the questions in the exercises for Class 12 Math Line

Shiksha's Class 12 Linear Programming, NCERT Solutions provide clear explanation for the concepts of feasible and infeasible regions;

• A feasible region is the common area on a graph that satisfies all the given constraints of a linear programming problem, including non-negativity conditions.
• An infeas

The Class 12 chapter 12 Linear Programming carries a weightage of 5 to 6 marks in the Class 12 CBSE Board Exams. Most of the times one long-answer question and some times 2-3 small answer type questions are asked from this chapter. Our NCERT Solutions will help students get full marks through the pr

Students can use Shiksa's NCERT Solutions for  Class 12 Linear Programming in PDF format;

• Open Shiksha's NCERT Solutions Link: Chapter 12 Linear Programming Solutions
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There are important topics in Class 12 Linear Programming which are asked in the CBSE 12 Boards. Class 12 Board questions are from important topics such as linear programming problems (LPP), graphical method of solving LPPs, feasible region, bounded/unbounded solutions, and corner point method. Clas