Question

A manufacturer can sell x items at the rate of \[\left( 330-x \right)\] each. The cost of producing items is Rs. \[\left( {{x}^{2}}+10x-12 \right)\] . How many items must be sold so that his profit is maximum?
(A) 80
(B) 20
(C) 70
(D) 50

Answer 1

Hint: The manufacturer has to sell x items. The cost of selling one item is Rs. \[\left( 330-x \right)\] . Now, calculate the cost of selling x items. It is given that the total cost for the production of x items is equal to Rs. \[\left( {{x}^{2}}+10x-12 \right)\] . Now, calculate the profit after deducting the total cost for the production of x items from the cost of selling x items. We have a function for profit. In calculus, we know that the slope of a function at its maxima is equal to zero. The slope of a function is its first derivative. Now, differentiate the function for the profit with respect to x and make it equal to zero. Then, solve it further and get the value of x.

Complete step-by-step answer:
According to the question, we have been given that a manufacturer can sell x items at the rate of \[\left( 330-x \right)\] each. The cost of producing items is Rs. \[\left( {{x}^{2}}+10x-12 \right)\] .
The cost of selling one item = Rs. \[\left( 330-x \right)\] ……………………………………..(1)
The total number of items = Rs. x …………………………………………(2)
The total cost of selling x items = Rs. \[x\left( 330-x \right)\] ………………………………….(3)
The total cost for the production of x items = Rs. \[\left( {{x}^{2}}+10x-12 \right)\] …………………………….(4)
With the help of selling price and the production cost, we can calculate the profit.
The profit for selling x items = Rs. \[x\left( 330-x \right)\] - Rs. \[\left( {{x}^{2}}+10x-12 \right)\] = Rs. \[\left( 330x-{{x}^{2}}-{{x}^{2}}-10x+12 \right)\] = Rs. \[\left( -2{{x}^{2}}+320x+12 \right)\] …………………………..…(5)
We have to find the number of items that must be sold so that his profit is maximum.
In calculus, we know that the slope of a function at its maxima is equal to zero.
From equation (5), we have the function of the profit.
We know that the slope of a function is its first derivative.

Now, on differentiating the function for the profit with respect to x and making it equal to zero, we get
\[\Rightarrow \dfrac{d\left( -2{{x}^{2}}+320x+12 \right)}{dx}=0\]
\[\Rightarrow \dfrac{d\left( -2{{x}^{2}} \right)}{dx}+\dfrac{d\left( 320x \right)}{dx}+\dfrac{d\left( 12 \right)}{dx}=0\] ……………………………….(6)
We know that \[\dfrac{d\left( a{{x}^{n}} \right)}{dx}=an{{x}^{n-1}}\] , \[\dfrac{d\left( ax \right)}{dx}=a\] , and \[\dfrac{d\left( \text{constant} \right)}{dx}=0\] …………………………………(7)
Now, from equation (6) and equation (7), we get
\[\begin{align}
  & \Rightarrow \left( -2 \right)2{{x}^{2-1}}+320+0=0 \\
 & \Rightarrow -4x+320=0 \\
 & \Rightarrow 4x=320 \\
 & \Rightarrow x=\dfrac{320}{40} \\
 & \Rightarrow x=80 \\
\end{align}\]
So, when \[x=80\] then, the function of the profit is maximum.
Since x is the total number of items sold, so the total number of items that must be sold so that his profit is maximum is 80.
So, the correct answer is “Option A”.

Note: In this question, one might take Rs. \[\left( 330-x \right)\] equal to the total cost of selling x items. This is wrong because the cost of selling one item is Rs. \[\left( 330-x \right)\] and the total number of items is x. Therefore, the cost of selling x items is Rs. \[x\left( 330-x \right)\] .