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 $2 x+y \leq 20, x+2 y \leq 12, x+3 y \leq 15, x \geq 0, y \geq 0$.
A firm has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ₹ 400 and each small van is ₹ 200. Not more than ₹ $3000$ is to be spent on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.
Let the firm has $x$ number of large vans and $y$ number of small vans. From the given information, we have following corresponding constraint table.
| Large vans ($x$) |
Small vans ($y$) | Maximum / Minimum | |
|---|---|---|---|
| Packages | 200 | 80 | 1200 |
| Cost | 400 | 200 | 3000 |
Thus, we see that objective function for minimum cost is $Z=400 x+200 y$. Subject to constraints
$$\begin{aligned} & 200 x+80 y \geq 1200 \quad \text { [package constraint] }\\ \Rightarrow \quad & 5 x+2 y \geq 30 \quad\text{.... (i)}\\ \text { and } \quad & 400 x+200 y \leq 3000 \quad \text { [cost constraint] }\\ \Rightarrow \quad & 2 x+y \leq 15 \quad\text{.... (ii)}\\ \text { and } \quad & x \leq y \text { [van constraint] ... (iii)}\\ \text { and } \quad & x \geq 0, y \geq 0 \text { [non-negative constraints] .... (iv) } \end{aligned}$$
Thus, required LPP to minimise cost is minimise $Z=400 x+200 y$, subject to $5 x+2 y \geq 30$.
$$\begin{aligned} 2 x+y & \leq 15 \\ x & \leq y \\ x & \geq 0, y \geq 0 \end{aligned}$$