Question

A company manufactures two types of hats, A and B. Market survey has indicated that the combined production level should not exceed 1200 hats per week and the demand for hats of type B is almost half of that for hats of type A. Further the production level of hats of type A can exceed three times the production of hats of type B by almost 600 units. The company makes a profit of Rs. 24 and Rs. 32 per hat of types A and B respectively. How many hats of each type should be produced weekly in order to maximize the profit?

Answer 1

Hint: Here the given question is a problem of linear programming, to solve this, we have to take \[x\] be a units of hat A and \[y\] be a units of hat B, next we have to write a mathematical expression using a statements in the given question and further plot a graph of LPP using a coordinate of each mathematical expression by the graph we can find easily the maximum profit.

Complete step by step solution:
Let’s make \[x\] be units of hat A and \[y\] be units of hat B.
Given,
Combine production level of hats A and B should not exceed 1200 has per week i.e.,
\[x + y \leqslant 1200\] ----(1)
Since, the production level of hats of type A can exceed three times the production of hats of type B by almost 600 units.
\[x - 3y \leqslant 600\] ----(2)
Also, demands for hats of type B is almost half of that for hats of type A, then
\[y \leqslant \dfrac{x}{2}\]
\[2y \leqslant x\]
Or
\[2y - x \leqslant 0\] ----(3)
As we know the count of an object cannot be negative, then
\[x \geqslant 0\], \[y \geqslant 0\]-----(4)
Given, profit of type \[A = 24\] and profit of type \[B = 32\], then
We need to find the maximize profit i.e., maximum of \[Z = 24x + 32y\]------(5)
Now, find the coordinates of each equation to plot the graph.
Now, consider equation (1)
\[ \Rightarrow x + y \leqslant 1200\]
Put, \[x = 0\] then
\[ \Rightarrow 0 + y = 1200\]
\[ \Rightarrow y = 1200\]
\[\therefore \left( {x,y} \right) = \left( {0,1200} \right)\]
Put, \[y = 0\] then
\[ \Rightarrow x + 0 = 1200\]
\[ \Rightarrow x = 1200\]
\[\therefore \left( {x,y} \right) = \left( {1200,0} \right)\]
The coordinates of the \[x + y \leqslant 1200\] are \[\left( {1200,0} \right)\] and \[\left( {0,1200} \right)\].
Now, consider equation (2)
\[ \Rightarrow x - 3y \leqslant 600\]
Put, \[x = 0\] then
\[ \Rightarrow 0 - 3y = 600\]
\[ \Rightarrow - 3y = 600\]
Divide both side by -3, then
\[ \Rightarrow y = - 200\]
\[\therefore \left( {x,y} \right) = \left( {0, - 200} \right)\]
Put, \[y = 0\] then
\[ \Rightarrow x - 3\left( 0 \right) = 600\]
\[ \Rightarrow x - 0 = 600\]
\[ \Rightarrow x = 600\]
\[\therefore \left( {x,y} \right) = \left( {600,0} \right)\]
The coordinates of the \[x - 3y \leqslant 600\] are \[\left( {0, - 200} \right)\] and \[\left( {600,0} \right)\].
Now, consider equation (3)
\[ \Rightarrow 2y - x \leqslant 0\]
Put, \[x = 400\] then
\[ \Rightarrow 2y - 400 = 0\]
\[ \Rightarrow 2y = 400\]
Divide both side by 2, then
\[ \Rightarrow y = 200\]
\[\therefore \left( {x,y} \right) = \left( {400,200} \right)\]
Put, \[y = 0\] then
\[ \Rightarrow 2\left( 0 \right) - x = 0\]
\[ \Rightarrow 0 - x = 0\]
\[ \Rightarrow x = 0\]
\[\therefore \left( {x,y} \right) = \left( {0,0} \right)\]
The coordinates of the \[2y - x \leqslant 0\] are \[\left( {400,200} \right)\] and \[\left( {0,0} \right)\].
Now, consider equation (4)
\[ \Rightarrow x \geqslant 0\] and \[y \geqslant 0\]
Now, plot the graph using all the obtained points from the above equation

In the graph, the shaded region represents the set of feasible solutions.
The coordinates of the corner points of the feasible region are \[O\left( {0,0} \right)\], \[A\left( {800,400} \right)\], \[B\left( {1050,150} \right)\] and \[C\left( {600,0} \right)\].
Now, consider the equation (5)
\[ \Rightarrow Z = 24x + 32y\]
The value of \[Z\] at \[O\left( {0,0} \right)\]
\[ \Rightarrow Z = 24\left( 0 \right) + 32\left( 0 \right)\]
\[\therefore Z = 0\]
The value of \[Z\] at \[A\left( {800,400} \right)\]
\[ \Rightarrow Z = 24\left( {800} \right) + 32\left( {400} \right)\]
\[ \Rightarrow Z = 19,200 + 12,800\]
\[\therefore Z = 32,000\]
The value of \[Z\] at \[B\left( {1050,150} \right)\]
\[ \Rightarrow Z = 24\left( {1050} \right) + 32\left( {150} \right)\]
\[ \Rightarrow Z = 25,200 + 4,800\]
\[\therefore Z = 30,000\]
The value of \[Z\] at \[C\left( {600,0} \right)\]
\[ \Rightarrow Z = 24\left( {600} \right) + 32\left( 0 \right)\]
\[ \Rightarrow Z = 14,400 + 0\]
\[\therefore Z = 14,400\]
Therefore, 800 units of hat A and 400 units of hat B should be produced weekly to get the maximum profit Rs. \[32,000\].
So, the correct answer is “Rs. \[32,000\]”.

Note: Linear programming is a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. Linear programming is a technique for the optimization of a linear function subject to linear equality and linear inequality constraints. Linear equation is an equation between two variables that gives a straight line when plotted on a graph remember to find feasible region Shade above the line for a “greater than” \[\left( {y > ory \geqslant } \right)\] or below the line for a “less than” \[\left( {y < ory \leqslant } \right)\].