Question

The total cost of producing x pocket ratio sets per day is Rs$\left( {\dfrac{1}{4}{x^2} + 35x + 25} \right)$
and the price at which they may be sold is Rs$\left( {50 - \dfrac{1}{2}x} \right)$. What should be the daily output to obtain a maximum total profit?
A. 5 sets
B. 10 sets
C. 15 sets
D. 20 sets

Answer 1

Hint: We know the cost of producing x product per day and price per set at which they can be sold. So, we find the profit function P by subtracting the selling price of x set and cost price of x sets. We get a function in x. We find the derivative of the function, equate it to 0 to see at what ‘x’, we get maximum profit.

Complete step by step solution:
Given cost of producing x pocket ratio sets per day$ = {\text{Rs}}\left( {\dfrac{1}{4}{x^2} + 35x + 25} \right)$
It is also given that,
Selling price per set$ = {\text{Rs}}\left( {{\text{50}} - \dfrac{1}{2}x} \right)$
Selling price of x sets$ = {\text{Rs}}\;x\left( {50 - \dfrac{1}{2}x} \right)$
 To find profit function, P we get it by subtracting the cost price of ‘x’ pocket radio set from selling price of x radio sets.
$Therefore, $ Profit function, P$ = x\left( {50 - \dfrac{1}{2}x} \right) - \left( {\dfrac{1}{4}{x^2} + 35x - 25} \right)$
$Therefore, {\text{P}}\;{\text{ = }}\left( {\dfrac{{ - 3}}{4}{x^2} + 15x - 25} \right)$
To check at which at x P attains its maximum, we find the derivative of P with respect to x.
i.e $\dfrac{{dp}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{ - 3}}{4}{x^2} + 15x - 25} \right)$
$\dfrac{{dp}}{{dx}} = \dfrac{{ - 3}}{2}x + 15$
Since, $\left( {\dfrac{d}{{dx}}\left( {a{x^n}} \right) = an{x^{n - 1}}} \right)$ where $a \in $ constant
 Now for P to be maximum
$\dfrac{{dp}}{{dx}} = 0$
$ \Rightarrow \dfrac{{ - 3}}{2}x + 15 = 0$
$Therefore, x = 10$
Also, $\dfrac{{{d^2}p}}{{d{x^2}}} = \dfrac{{ - 3}}{2} < 0$
Hence P attains maximum at $x = 10$
i.e., For maximum profit the daily profit output must be 10 radio sets

Correct option B 10 sets.

Note: One must remember the basic fact that profit equals the difference between the selling price and cost price
i.e., Profit $=$selling price$-$1 at cost price
To check when and what point a function attains maximum or minimum, we first find the derivative of the function, equate it to 0 to get at what point function attains maximum or minimum. We then find a double derivative and see if it is less than 0 then function attains a maximum and if it is greater than 0, then function attains minimum. We do it by substituting the value in the double derivative, we calculated.