$\begin{array}{1 1} 32 \\ 30 \\ 0 \\ 33 \end{array} $

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- First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
- One we graphically plot the area bounded by the constraints, it’s easy to see which points satisfy all constraints. This common region determined by all the constraints including non-negative constraints of a linear programming problem is called the $\textbf{Feasible Region (or solution region).}$
- Now, any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an $\textbf{Optimal Solution}$. We see that every point in the feasible region satisfies all the constraints, and there are infinitely many points.
- Since we know from theory that the optimal value must occur at a corner point (vertex) of the feasible region, calculate the objective function values associated with the coordinates of all the extreme points. This method is called the $\textbf{Corner Point Method}$
- If the feasible region is bounded (if it can be enclosed), the point with the best objective function value is the best optimal solution. If the feasible region is unbounded (means that the feasible region does extend indefinitely in any direction), the then a maximum or a minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of the feasible region, which can be calculated

Let x be the number of pedestal lamps and y be the number of wooden shades that we can make. Our problem is to maximize x and y and the maximum profit at a given capacity

Clearly, x, y ≥ 0. Let us construct the following table from the given data

Pedestal Lamps (x) | Wooden Shades (y) | Time available | |

Grinding Machine (h) | 2 | 1 | 12h |

Sprayer (h) | 3 | 2 | 20h |

Profits (Rs.) | 5 | 3 |

We have the following constraints: 2x+y $\leq$ 12 and 3x+2y$\leq$20

The profit on pedestal lamps s Rs. 5 and on wooden lampshades is Rs. 3. We need to maximize the profits, i.e. maximize 5x + 3y, given the above constraints.

$\textbf{Plotting the constraints}$:

Plot the straight lines 2x + y = 12 and 3x + 2y = 20

First draw the graph of the line 2x + y = 12

If x = 0, y = 12 and if y = 0, x = 6. So, this is a straight line between (0,12) and (6,0).

At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 0. So the area associated with this inequality is bounded towards the origin.

Similarly, draw the graph of the line 3x + 2y = 20.

If x = 0, y = 10 and if y =0, x = 20/3. So, this is a straight line between (0,10) and (20/3,0).

At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 0. So the area associated with this inequality is bounded towards the origin.

$\textbf{Finding the feasible region}$:

We can see that the feasible region is bounded and in the first quadrant.

On solving the equations 2x + y = 12 and 3x + 2y = 20, we get,

3x + 2 (12-2x) = 20 $\to$3x + 24 – 4x = 20 $\to$x = 4.

If x = 4, y = 12 – 2x4 = 4.

$\Rightarrow x = 4, y = 4 $

Therefore the feasible region has the corner points (0,0), (0,10), (4,4), (6,0) as shown in the figure.

$\textbf{Solving the objective function using the corner point method}$

The values of Z at the corner points are calculated as follows:

Corner Point | Z = 5x+3y |

O (0,0) | 0 |

C (0,10) | 30 |

E (4,4) | 32 (Max Value) |

B (6.0) | 30 |

$\textbf{The maximum profit we can make is Rs. 32, which involves making 4 pedestal lamps and 4 wooden shades. }$

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