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# A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?

Toolbox:
• 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 nuts and y be the number of bolts that we can make. Our problem is to maximize x and y if the factory worked at capacity of 12h / day. We also have to find out the maximum profit at this capacity.
Clearly, x, y ≥ 0. Let us construct the following table from the given data
 Nuts (x) Bolts (y) # hrs avail Machine A (h) 1 3 12 Machine B (h) 3 1 12 Profits (Rs.) 17.5 7
Since we have only 12h of machine time, we have the following constraints:
x + 3y $\leq$ 12
3x + y $\leq$ 12
The profits on the nuts is Rs. 17.5 and on the bolts is Rs. 7. We need to maximize the profits, i.e. maximize 17.5x + 7y, given the above constraints.
$\textbf{Plotting the constraints}$:
Plot the straight lines x + 3y = 12 and 3x + y = 12.
First draw the graph of the line x + 3y = 12
If x = 0, y = 4 and if y = 0, x = 12. So, this is a straight line between (0,4) and (12,0).
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 + y = 12.
If x = 0, y = 12 and if y =0, x = 4. So, this is a straight line between (0,12) and (4,0).
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 x+3y = 12 and 3x + y = 12, we get,
x + 3 (12-3x) = 12 $\to$x + 36 – 9x = 12 $\to$-8x = -24 $\to$x = 3.
If x = 3 then y = 12 – 9 = 3.
$\Rightarrow x = 3, y = 3$
Therefore the feasible region has the corner points (0,0), (0,4), (3,3), (4,0) 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 = 17.5x + 7y (0,0) 0 (0,4) 28 (3,3) 73.5 (Max Value) (4,0) 70
$\textbf{ The maximum profit we can make is Rs. 73.5, which involves making 3 packages of nuts and bolts each. }$

edited Apr 18, 2013