Question

A manufacturer can sell $x$ computers at a price of Rs. $(330 - x)$ each. The cost of production $x$ computers is Rs. $({x^2} + 10x - 12).$ Determine the number of computers to be sold so that he can make maximising output.

Answer 1

Hint : First of all we will frame the equation for the profit by the given selling and cost price. Then we will find its first and second order derivatives and then will find the value which maximizes the profit.

Complete step-by-step answer :
Given that the selling price of item is $= Rs.{\text{ (330 - x)}}$
Therefore, the selling price of x item is $= Rs.{\text{ x(330 - x)}}$
Also, given that the cost price of “x” items $= ({x^2} + 10x - 12).$
According to the formula –
Profit $=$ Selling price – cost price
Place values in the above relation –
$\Rightarrow \Pr ofit = (330x - {x^2}) - ({x^2} + 10x - 12)$
Simplify the above equation. When you open the brackets and there is a negative sign outside it, then the sign of all the terms inside the bracket changes. Positive term becomes negative and negative term becomes positive.
$\Rightarrow \Pr ofit = 330x - {x^2} - {x^2} - 10x + 12$
Make pair of like terms in the above equation-
$\Rightarrow \Pr ofit = \underline {330x - 10x} - \underline {{x^2} - {x^2}} + 12$
Simplify the above equations –
$\Rightarrow \Pr ofit = 320x - 2{x^2} + 12$
The above equation can be re-written as –
$\Rightarrow \operatorname{P} = - 2{x^2} + 320x + 12$
Differentiate above equation with respect to “x”
$\Rightarrow \dfrac{{d\operatorname{P} }}{{dx}} = \dfrac{d}{{dx}}\left( { - 2{x^2} + 320x + 12} \right)$
When derivative is outside the bracket then we have to apply derivative inside the bracket to all the terms.
$\Rightarrow \dfrac{{d\operatorname{P} }}{{dx}} = \left( {\dfrac{d}{{dx}}\left( { - 2{x^2}} \right) + \dfrac{d}{{dx}}\left( {320x} \right) + \dfrac{d}{{dx}}\left( {12} \right)} \right)$
Now, using the identity $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$ and also, $\dfrac{{d(c)}}{{dx}} = 0$ derivative of constant numbers is always zero.
$\Rightarrow \dfrac{{d\operatorname{P} }}{{dx}} = - 4x + 320$ .... (A)
Again differentiating the above equation with respect to “x”
$\Rightarrow \dfrac{{{d^2}\operatorname{P} }}{{d{x^2}}} = - 4$
The above value is less than zero.
So, to maximize the profit, take equation (A) equal to zero.
$\Rightarrow \dfrac{{d\operatorname{P} }}{{dx}} = 0$
$0 = - 4x + 320$
Move term with variable on the left hand side of the equation. When you move any term from one side to another, the sign is also changed. Positive terms become negative and vice-versa.
$\Rightarrow 4x = 320$
When the term multiplicative on one side is moved to the opposite side, then it goes to the denominator.
$\Rightarrow x = \dfrac{{320}}{4}$
$\Rightarrow x = \dfrac{{80 \times 4}}{4}$
Common factors from the numerator and the denominator cancel each other.
$\Rightarrow x = 80$
Thus, $80$ items must be sold for maximizing the profit.
So, the correct answer is “x=80”.

Note : Always remember when you move any term from one side to another, the sign of the term also changes. Positive terms become negative and negative terms become positive. When you open the brackets and there is a negative sign outside it, then sign of all the terms inside the bracket changes but when there is a positive sign outside the bracket then there is no change in the signs of the terms inside the bracket. Be careful about the sign.