Examples (Revised) - Chapter 12 - Linear Programming - Ncert Solutions class 12 - Maths

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Chapter 12: Linear Programming - NCERT Solutions Class 12 Maths

Example 1

Solve the following linear programming problem graphically:
$
\text { Maximise } \mathrm{Z}=4 x+y
$
subject to the constraints:
$
\begin{aligned}
x+y & \leq 50 \\
3 x+y & \leq 90 \\
x \geq 0, y & \geq 0
\end{aligned}
$

Solution

The shaded region in Fig 12.2 is the feasible region determined by the system of constraints (2) to (4). We observe that the feasible region OABC is bounded. So, we now use Corner Point Method to determine the maximum value of $\mathrm{Z}$.

The coordinates of the corner points $\mathrm{O}, \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $(0,0),(30,0),(20,30)$ and $(0,50)$ respectively. Now we evaluate $Z$ at each corner point.

Hence, maximum value of $\mathrm{Z}$ is 120 at the point $(30,0)$.
Example 2

Solve the following linear programming problem graphically:
Minimise $\mathrm{Z}=200 x+500 y$
subject to the constraints:
$
\begin{aligned}
x+2 y & \geq 10 \\
3 x+4 y & \leq 24 \\
x \geq 0, y & \geq 0
\end{aligned}
$

Solution

The shaded region in Fig 12.3 is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner points

A, B and C are $(0,5),(4,3)$ and $(0,6)$ respectively. Now we evaluate $Z=200 x+500 y$ at these points.
Hence, minimum value of $\mathrm{Z}$ is 2300 attained at the point $(4,3)$
Example 3

Solve the following problem graphically:
$
\begin{aligned}
& \text { Minimise and Maximise } \mathrm{Z}=3 x+9 y \\
& \text { subject to the constraints: } \quad x+3 y \leq 60 \\
& x+y \geq 10 \\
& x \leq y \\
& x \geq 0, y \geq 0 \\
&
\end{aligned}
$

Solution

First of all, let us graph the feasible region of the system of linear inequalities (2) to (5). The feasible region $\mathrm{ABCD}$ is shown in the Fig 12.4. Note that the region is bounded. The coordinates of the corner points A, B, C and D are $(0,10),(5,5),(15,15)$ and $(0,20)$ respectively.

We now find the minimum and maximum value of $\mathrm{Z}$. From the table, we find that the minimum value of $\mathrm{Z}$ is 60 at the point $\mathrm{B}(5,5)$ of the feasible region.

The maximum value of $Z$ on the feasible region occurs at the two corner points $C(15,15)$ and $D(0,20)$ and it is 180 in each case.

Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points $C$ and D, i.e. the both points produce same maximum value 180 . In such cases, you can see that every point on the line segment CD joining the two corner points $\mathrm{C}$ and $\mathrm{D}$ also give the same maximum value. Same is also true in the case if the two points produce same minimum value.

Example 4

Determine graphically the minimum value of the objective function
$
\mathrm{Z}=-50 x+20 y
$
subject to the constraints:
$
\begin{aligned}
& 2 x-y \geq-5 \\
& 3 x+y \geq 3 \\
& 2 x-3 y \leq 12 \\
& x \geq 0, y \geq 0
\end{aligned}
$

Solution

First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the Fig 12.5. Observe that the feasible region is unbounded.
We now evaluate $\mathrm{Z}$ at the corner points.

From this table, we find that -300 is the smallest value of $\mathrm{Z}$ at the corner point $(6,0)$. Can we say that minimum value of $Z$ is -300 ? Note that if the region would have been bounded, this smallest value of $Z$ is the minimum value of $Z$ (Theorem 2). But here we see that the feasible region is unbounded. Therefore, -300 may or may not be the minimum value of $\mathrm{Z}$. To decide this issue, we graph the inequality
i.e.,
$
\begin{aligned}
& -50 x+20 y<-300 \text { (see Step 3(ii) of corner Point Method.) } \\
& -5 x+2 y<-30
\end{aligned}
$
and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then -300 will not be the minimum value of $Z$. Otherwise, -300 will be the minimum value of $\mathrm{Z}$.

As shown in the Fig 12.5, it has common points. Therefore, $Z=-50 x+20 y$ has no minimum value subject to the given constraints.

In the above example, can you say whether $z=-50 x+20 y$ has the maximum value 100 at $(0,5)$ ? For this, check whether the graph of $-50 x+20 y>100$ has points in common with the feasible region. (Why?)

Example 5

Minimise $Z=3 x+2 y$
subject to the constraints:
$
\begin{aligned}
x+y & \geq 8 \\
3 x+5 y & \leq 15 \\
x \geq 0, y & \geq 0
\end{aligned}
$

Solution

Let us graph the inequalities (1) to (3) (Fig 12.6). Is there any feasible region? Why is so?

From Fig 12.6, you can see that there is no point satisfying all the constraints simultaneously. Thus, the problem is having no feasible region and hence no feasible solution.

Remarks From the examples which we have discussed so far, we notice some general features of linear programming problems:
(i) The feasible region is always a convex region.
(ii) The maximum (or minimum) solution of the objective function occurs at the vertex (corner) of the feasible region. If two corner points produce the same maximum (or minimum) value of the objective function, then every point on the line segment joining these points will also give the same maximum (or minimum) value.