Question

The profit function, in rupees, of a firm selling $x$ items $\left( x\ge 0 \right)$ per week is given by $P\left( x \right)=-3500+\left( 400-x \right)x$. How many items should the firm sell so that the firm has maximum profit?
A. 400
B. 300
C. 200
D. 100

Answer 1

Hint:
Here, in the given question, we are given that the profit function of a firm selling $x$ items where $x\ge 0$, per week, is $P\left( x \right)=-3500+\left( 400-x \right)x$ (in rupees) and we need to find the number of items the firm should sell so that the firm will have maximum profit. Profit maximisation is the technique used by corporations and enterprises to develop methods to increase earnings while reducing costs. Every company should have it as a primary goal because it is essential to their development.

Formula used:
The derivative(y') of a constant is 0
$y = C$ then $y' = 0$
$\dfrac{d}{dx}{x^n}=nx^{n-1}$

Complete step-by-step solution:
We have, $P\left( x \right)=-3500+\left( 400-x \right)x$
$\Rightarrow P\left( x \right)=-3500+400x-{{x}^{2}}$
Now, we will differentiate the above-written equation w.r.t. $x$
$\Rightarrow P'\left( x \right)=400-2x$
For the maximum value of $x$,
$P'\left( x \right)=0$.
$\therefore 0=400-2x$
$\Rightarrow 2x=400$
Divide the equation by $2$
$\Rightarrow x=200$
On again differentiating w.r.t. $x$,
$P''\left( x \right)=-2<0$
Here,
$P''\left( x \right)<0$
It means $x=200$ is the point of the local maximum.
So $P\left( x \right)$ is maximum at $x=200$.
Hence, the firm should sell $200$ items to get the maximum profit.

So the correct answer is option (B)

Note: Remember that the first derivative test is for local maxima and minima. For function $f\left( x \right)$ which is differentiable at $x=a$, is maximum or minimum if $f'\left( a \right)=0$. Second order derivative or higher order derivative enables us to find the points of local maxima or local minima.