Question

A firm makes items $A$ and $B$ and the total number of items it can make in a day is $24$ It takes one hour to make an item of $A$ and only half an hour to make an item of $B$. The maximum time available per day is $16$ hours. The profit on an item of $A$ is $Rs.300$ and on one item of $B$ is $Rs.160$. How many times of each type should be produced to maximize the profit? Solve the problem graphically.

Answer 1

The given problem can be solved as a linear programming problem.

Let the number of items of A per day be $x$.

Let the number of items of B per day be $y$.

Number of items of A and B manufactured per day is a max. of 24.

Thus $x+y\le 24...\left(i\right)$

1 item of A requires 1 hour to manufacture.

Hence $x$ items of A require $x$ hours.

Similarly, $y$ items of B require $y/2$ hours.

The total manufacturing time available per day=16 hours.

Hence,

$x+\frac{y}{2}\le 16...\left(ii\right)$

Now the total profit $Z$ is the sum of profit on A and profit on B.

Profit per item of A=300 and per item of B=160.

Thus,

Maximize $Z=300x+160y$.

The above graph is obtained.

The corner points are A(0,24), B(8,16) and C(16,0).

The value of Z at these points is,

At $A(0,24)$

$Z=300\cdot 0+160\cdot 24=3840$

At $B(8,16)$

$Z=300\cdot 8+160\cdot 16=4960$

At $C(16,0)$

$Z=300\cdot 16+160\cdot 0=4800$

Hence the profit is maximum at $B(8,16)$.

Thus the units of A to be manufactured is 8 and the units of B to be manufactured is 16.