Question

A manufacturer produces two products $A$ and $B$. Product $A$ fetches him a profit of Rs. $24$ and product $B$ fetches him a profit of Rs. $14$. It takes $15$ minutes to manufacture one unit of product $A$ and $5$ minutes to manufacture one unit of product $B$. There is a limit of $30$ units to the quantity of $B$ of manufactured due to a constraint on the availability of materials. The minimum profit that the company decides to make is Rs. $1000$. The workers work for a maximum of $10$ hours a day. Find the maximum daily profit.

• A. Rs. $1020$
• B. Rs. $1040$
• C. Rs. $1140$
• D. Rs. $1240$

Answer 1

Answer: (C)

Rs. $1140$

Given, The profit from the product $A$ is $Rs.24$ and

The profit from the product $B$ is $Rs.14$

The maximum number of units of $B$ manufactured in a day is $30$

i.e., $y\le 30$ --------- (1)

Let the number of units of $A$ manufactured in a day be $x$

Therefore the profit for a day from the product $A$ is $24x$

Let the number of units of $B$ manufactured in a day be $y$

Therefore the profit for a day from the product $B$ is $14y$

The minimum profit for a day is $Rs.1000$

Therefore, the total profit from products $A$ and $B$ should be more than $Rs.1000$

i.e., $24x+14y\ge 1000$ --------- (2)

Given, time taken to manufacture one product of $A$ is $15min$ and

time taken to manufacture one product of $B$ is $5min$

Therefore, time taken to manufacture $x$ products of $A$ is $15xmin$ and

time taken to manufacture $y$ products of $B$ is $5ymin$

In a day, a worker works for a maximum of $10hrs=600min$

Therefore, the time taken to manufacture products $A$ and $B$ should be less than $600min$

I.e., $15x+5y\le 600$ --------- (3)

The total profit from products $A$ and $B$ is $P=24x+14y$

In the above figure, the blue shaded region is the feasible region with three corner points.$(\frac{290}{12},30),(30,30),(\frac{340}{9},\frac{20}{3})$

$(\frac{290}{12},30)$ is the point where $24x+14y=1000$ intersects $y=30$

I.e., substituting $y=30⟹24x+14\ast 30=1000⟹x=\frac{1000-420}{24}⟹x=\frac{290}{12}$

$(30,30)$ is the point where $15x+5y=600$ intersects $y=30$

I.e., substituting $y=30⟹15x+5\ast 30=600⟹x=\frac{600-150}{15}⟹x=30$

$(\frac{340}{9},\frac{20}{3})$ is the point where $24x+14y=1000$ intersects $15x+5y=600$

I.e., solving the two equations, we get $x=\frac{340}{9}$and $y=\frac{20}{3}$

Now substituting the corner points the profit equation,

substituting $(\frac{290}{12},30)⟹P=24\ast \frac{290}{12}+14\ast 30=1000$

substituting $(30,30)⟹P=24\ast 30+14\ast 30=1140$

substituting $(\frac{340}{9},\frac{20}{3})⟹P=24\ast \frac{340}{9}+14\ast \frac{20}{3}=1000$

$1140$ is the maximum profit