0
$\begingroup$

Here is the given question: A toy manufacturer produces two types of dolls; a basic version doll $A$ and a deluxe version doll $B$. Each doll of type $B$ takes twice as long to produce as one doll of type $A$. The company have time to make a maximum of $2000$ dolls of type $A$ per day, the supply of plastic is sufficient to produce $1500$ dolls per day and each type requires equal amount of it. The deluxe version, i.e type $B$ requires a fancy dress of which there are only $600$ available per day. If the company makes a profit of $₹ 3$ and $₹ 5$ per doll, respectively, on doll $A$ and doll $B$; how many of type should be produced per day in order to maximize the profit?

Now, in the solution the equation involving the time required and available is given as: $x + 2y ≤ 2000$ (where $x$ dolls of type $A$ and $y$ dolls of type $B$ are produced per day to maximize the profit.) But shouldn't the equation be $2x + y ≤ 2000$ OR $x + (1/2)y ≤ 2000$ ? Please explain how to interpret the problem correctly. In my logic, since type B dolls takes twice the time to produce compared to doll $A$, only half of them could be produced in a given time as compared to $A$.

$\endgroup$
1
$\begingroup$

I suppose that, implicitly, $x$ is the quantity of doll $A$ which is produced, and $y$ is the quantity of doll $B$ which is produced (on a given day). Suppose that one of each doll type is made. Then, in terms of time units it takes to make doll $A$, you have one time unit from the one doll $A$ produced and two time units from the one doll $B$ that is produced -- that is, it takes $x + 2y$ time units (with $x = 1$ and $y = 1$ in our example). Does this make sense?

$\endgroup$
  • $\begingroup$ okay..that part is clarified..but now can't relate with "The company have time to make a maximum of 2000 dolls of type A per day" $\endgroup$ – Raunak Bag Dec 31 '15 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.