Question

A company sells packets of biscuits each day at ₹ 10 a packet. The cost of manufacturing these packets is ₹ 5 per packet plus a fixed daily overhead cost of ₹ 700. What will be the profit function?

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

A company sells packets of biscuits each day at ₹ 10 a packet.

Let assume that company manufacturers 'x' packets.

So,

Revenue function, R(x)

$\rm \implies\:\boxed{ \tt{ \: R(x) = 10x \: }} \: - - - - (1)$

Further given that,

The cost of manufacturing these packets is ₹ 5 per packet plus a fixed daily overhead cost of ₹ 700.

We know,

↝ Cost function C(x) = Variable Cost + Fixed Cost

So,

$\rm \implies\:\boxed{ \tt{ \: C(x) = 700 + 5x \: }} \: - - - - (2)$

Now, We know that,

↝ Profit function, P(x) is given by

$\red{\rm :\longmapsto\: P(x) \: = \: R(x) \: - \: C(x) \: }$

On substituting the values of R(x) and C(x), we get

$\rm :\longmapsto\:P(x) \: = \: 10x - (700 + 5x)$

$\rm :\longmapsto\:P(x) \: = \: 10x - 700 - 5x$

$\rm \implies\:\boxed{ \tt{ \: P(x) = 5x - 700 \: }}$

Additional Information :-

1. Revenue function is

$\boxed{ \tt{ \: R(x) \: = \: p \: x}}$

2. Average Revenue function is

$\boxed{ \tt{ \: AR(x) = \frac{R(x)}{x} \: }}$

3. Marginal Revenue function is

$\boxed{ \tt{ \: MR(x) = \dfrac{d}{dx}R(x) \: }}$

4. Average Cost function is

$\boxed{ \tt{ \: AC(x) = \frac{C(x)}{x} \: }}$

5. Marginal Cost function is

$\boxed{ \tt{ \: MC(x) = \dfrac{d}{dx}C(x) \: }}$

Answer 2

Step-by-step explanation:

S.P=₹10x i.e R(x)

C(x)=5x+700

ACC. to formula

P(x)=R(x)-C(x)

=10x-(5x+700)

=10x-5x-700

P(x) i.e profit fn =5x-700