Question

A manufacturer has employed $5$ skilled men and $10$ semi-skilled men and makes two models $A$ and $B$ of an article. The making of one item of model $A$ requires $2$ hours work by a skilled men and $2$ hours work by a semi-skilled man. One item of model $B$ requires $1$ hour by a skilled man and $3$ hours by a semi-skilled man. No man is expected to work more than $8$ hours per day. The manufacturer's profit on an item of model $A$ is Rs.$15$ and on an item of model $B$ is Rs.$10,$ then the total number of items of each model should be made per day in order to maximize daily profit.
$[$where, $n\left(A\right)=$ number of items of artical $A$ and $n\left(B\right)=$ number of items of artical $A]$

• A. $n\left(A\right)=10,n\left(B\right)=15$
• B. $n\left(A\right)=15,n\left(B\right)=10$
• C. $n\left(A\right)=10,n\left(B\right)=20$
• D. $n\left(A\right)=20,n\left(B\right)=10$

Answer 1

The correct option is C $n\left(A\right)=10,n\left(B\right)=20$

Given that total available hours for skilled men $=8\times 5=40$
and total available hours for semi-skilled men $=8\times 10=80$
Let $x$ be the number of items produced of model $A$ and $y$ be the number of items produced of model $B.$

Let $Z$ be the maximizing function.
$\text{Then}Z=15x+10y$
subject to the constraints
$2x+y\le 40\left(\text{skilled men work time constraint}\right)2x+3y\le 80\left(\text{semi-skilled men work time constraint}\right)x\ge 0,y\ge 0\left(\text{non-negative constraints}\right)$

Plotting the graphs form the above constraints, we have

From the above graph, we get

Corner points	$Z=15x+10y$
$(0,26.667)$	$266.67$
$(10,20)$	$350$
$(0,0)$	$0$
$(20,0)$	$300$

As we can see that the maximum value of $Z$ occurs at $(10,20).$
So, manufacturer should produce $10$ items of model $A$ and $20$ items of model $B$ in order to maximize the profit.
The maximum profit is Rs.$350$