The feasible region for a LPP is shown in following figure. Find the minimum value of $Z=11 x+7 y$.
From the figure, it is clear that feasible region is bounded with coordinates of corner points as $(0,3),(3,2)$ and $(0,5)$. Here, $Z=11 x+7 y$.
$$\begin{aligned} & \because \quad x+3 y=9 \text { and } x+y=5 \\ & \Rightarrow \quad 2 y=4 \\ & \therefore \quad y=2 \text { and } x=3 \end{aligned}$$
So, intersection points of $x+y=5$ and $x+3 y=9$ is $(3,2)$.
| Corner points | Corresponding value of Z |
|---|---|
| $(0,3)$ | 21 $\leftarrow$ Minimum |
| $(3,2)$ | 47 |
| $(0,5)$ | 35 |
Hence, the minimum value of Z is 21 at (0, 3).
The feasible region for a LPP is shown in following figure. Find the maximum value of Z.
It is clear that Z is maximum at (3, 2) and its maximum value is 47.
The feasible region for a LPP is shown in the following figure. Evaluate $Z=4 x+y$ at each of the corner points of this region. Find the minimum value of $Z$, if it exists.
From the shaded region, it is clear that feasible region is unbounded with the corner points $A(4,0), B(2,1)$ and $C(0,3)$
Also, we have $$Z=4 x+y$$
[since, $x+2 y=4$ and $x+y=3 \Rightarrow y=1$ and $x=2$ ]
| Corner points | Corresponding value of Z |
|---|---|
| $(4,0)$ | 16 |
| $(2,1)$ | 9 |
| $(0,3)$ | 3 $\leftarrow$ Minimum |
Now, we see that 3 is the smallest value of $Z$ at the corner point $(0,3)$. Note that here we see that, the region is unbounded, therefore 3 may or may not be the minimum value of $Z$.
To decide this issue, we graph the inequality $4 x+y< 3$ and check whether the resulting open half plan has no point in common with feasible region otherwise, $Z$ has no minimum value.
From the shown graph above, it is clear that there is no point in common with feasible region and hence $Z$ has minimum value 3 at $(0,3)$.
In following figure, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of $Z=x+2 y$.
From the shaded bounded region, it is clear that the coordinates of corner points are $\left(\frac{3}{13}, \frac{24}{13}\right),\left(\frac{18}{7}, \frac{2}{7}\right),\left(\frac{7}{2}, \frac{3}{4}\right)$ and $\left(\frac{3}{2}, \frac{15}{4}\right)$.
Also, we have to determine maximum and minimum value of $Z=x+2y$.
| Corner points | Corresponding value of Z |
|---|---|
| $\left(\frac{3}{13}, \frac{24}{13}\right)$ | $\frac{3}{13}+\frac{48}{13}=\frac{51}{13}=3 \frac{12}{13}$ |
| $\left(\frac{18}{7}, \frac{2}{7}\right)$ | $\frac{18}{7}+\frac{4}{7}=\frac{22}{7}=3 \frac{1}{7}$ Minimum |
| $\left(\frac{7}{2}, \frac{3}{4}\right)$ | $\frac{7}{2}+\frac{6}{4}=\frac{20}{4}=5$ |
| $\left(\frac{3}{2}, \frac{15}{4}\right)$ | $\frac{3}{2}+\frac{30}{4}=\frac{36}{4}=9$ Maximum |
Hence, the maximum and minimum values of $Z$ are 9 and $3 \frac{1}{7}$, respectively.
A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits $A$ and $B$. Type $A$ requires 20 resistors, 10 transistors and 10 capacitors. Type $B$ requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type $A$ circuit is ₹ 50 and that on type $B$ circuit is ₹ 60 , formulate this problem as a LPP, so that the manufacturer can maximise his profit.
Let the manufacturer produces $x$ units of type $A$ circuits and $y$ units of type $B$ circuits. Form the given information, we have following corresponding constraint table.
| Type A ($x$) |
Type B ($y$) | Maximum stock |
|
|---|---|---|---|
| Resistors | 20 | 10 | 200 |
| Transistors | 10 | 20 | 120 |
| Capacitors | 10 | 30 | 150 |
| Profit | ₹ 50 | ₹ 60 |
Thus, we see that total profit $Z=50 x+60 y$ (in ₹).
Now, we have the following mathematical model for the given problem.
Maximise $$Z=50 x+60 y\quad\text{.... (i)}$$
Subject to the constraints.
$$\begin{aligned} 20 x+10 y & \leq 200 \quad\text{[resistors constraint]}\\ \Rightarrow\quad 2 x+y & \leq 20 \quad \text{.... (ii)}\\ \text{and}\quad 10 x+20 y & \leq 120 \quad\text{[transistor constraint]}\\ \Rightarrow\quad x+2 y & \leq 12 \quad\text{.... (iii)}\\ \text{and}\quad 10 x+30 y & \leq 150 \quad\text{[capacitor constraint]}\\ \Rightarrow\quad x+3 y & \leq 15 \quad\text{.... (iv)}\\ \text{and}\quad x \geq 0, y & \geq 0\quad\text{[non-negative constraint] .... (v)} \end{aligned}$$
So, maximise $Z=50 x+60 y$, subject to $2 x+y \leq 20, x+2 y \leq 12, x+3 y \leq 15, x \geq 0, y \geq 0$.