Question

A company produces two types of tables, ${{T}_{1}}$ and ${{T}_{2}}$. It takes 2 hours to produce the parts of one unit of ${{T}_{1}}$, 1 hour to assemble and 2 hours polish. It takes 4 hours to produce parts of one unit of ${{T}_{2}}$, 2.5 hour to assemble and 1.5 hours to polish. Per month, 7000 hours are available for producing the parts, 4000 hours for assembling the parts and 5500 hours for polishing the tables. The profit per unit of ${{T}_{1}}$ is 90 and per unit of ${{T}_{2}}$ is 110. How many of each type of tables should be produced in order to maximize the total monthly profit.

Answer 1

Hint: To solve this question, we will assume variables for number of ${{T}_{1}}$ tables produced and number of ${{T}_{2}}$ tables produced. Then, the total available hours for each process is given to us. The underlying principle to solve this problem is that the profit will be maximised if we use the available hours to the fullest. With the given information we will form linear equations in two variables and solve them so that hours are used to the maximum. Since, three linear equations will be formed, we will have two sets of solutions, the set which gives maximum profit will be chosen.

Complete step-by-step answer:
Let the number of tables of type ${{T}_{1}}$ be x and the number of tables of type ${{T}_{2}}$ be y.
Now, we know that the production of parts one unit of ${{T}_{1}}$ takes 2 hours and that of ${{T}_{2}}$ takes 4 hours and the total available hours for production of parts in one month is 7000 hours.
Therefore, if total hours required to produce parts of ${{T}_{1}}$ will be the product of number of ${{T}_{1}}$ tables and number of hours required to produce parts of one unit.
$\Rightarrow 2\times x$
And the total hours required to produce parts of ${{T}_{2}}$ will be the product of the number of ${{T}_{2}}$ tables and number of hours required to produce parts of one unit.
$\Rightarrow 4\times y$
Sum of hours to produce parts of x units of ${{T}_{1}}$ and that of y units of ${{T}_{2}}$ must be less than total hours available for producing parts.
$\Rightarrow 2x+4y\le 7000..................................\left( 1 \right)$
Similarly, we know that the number of hours to assemble one unit of ${{T}_{1}}$ is 1 hour and number of hours to assemble one unit of ${{T}_{2}}$ is 2.5 hours.
The sum of time required to assemble x units of ${{T}_{1}}$ and that of y units of ${{T}_{2}}$ must be less than total hours available for assembling.
$\Rightarrow x+2.5y\le 4000....................................\left( 2 \right)$
In the similar way, we know that number of hours to polish one unit of ${{T}_{1}}$ is 2 hours and number of hours to polish one unit of ${{T}_{2}}$ is 1.5 hours.
The sum of time required to polish x units of ${{T}_{1}}$ and that of y units of ${{T}_{2}}$ must be less than total hours available for polishing.
$\Rightarrow 2x+1.5y\le 5500.....................................\left( 3 \right)$
We know that the profit per unit of ${{T}_{1}}$ is 90 and that of ${{T}_{2}}$ is 110. Therefore, the total profit if x units of ${{T}_{1}}$ and y units of ${{T}_{2}}$ are produced will be
$90\left( x \right)+110\left( y \right)$
Thus, now we’ve got three linear equations.
First, we will solve equation (1) and (2). This means that the company is using all the hours available for producing and all the hours available for assembling.
We will get the solutions of the equations as x = 1500 and y = 1000.
Thus, if the company completely uses available hours for production and assembling, it can produce 1500 units of ${{T}_{1}}$ and 1000 units of ${{T}_{2}}$.
Therefore, profit will be
$90\left( 1500 \right)+110\left( 1000 \right)=245000$
Now, we will solve equation (1) and (3). This means that the company is using all the hours available for producing and all the hours available for polishing.
We will get the solutions of the equations as x = 2300 and y = 600.
Thus, if the company completely uses available hours for production and polishing, it can produce 2300 units of ${{T}_{1}}$ and 600 units of ${{T}_{2}}$.
Therefore, profit will be
$90\left( 2300 \right)+110\left( 600 \right)=273000$
Thus, it is clear that if the company produces 2300 units of type ${{T}_{1}}$ and 600 units of type ${{T}_{2}}$, the profit will be maximum.

Note: Students are free to use any method to solve the linear equations in two variables as the solution remains the same. Recommended method is to use elimination method. Because it's easy to understand and solve.