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

& \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{aligned}$$ So, when $$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)$$ .