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A Firm Has the Cost Function C = X^3/3 - 7x^2 + 111x + 50 And Demand Function X = 100 – P. Write the Total Revenue Function in Terms of X. - Mathematics

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Sum

A firm has the cost function `C = x^3/3 - 7x^2 + 111x + 50`  and demand function x = 100 – p.
(i) Write the total revenue function in terms of x.
(ii) Formulate the total profit function P in terms of x.
(iii) Find the profit-maximizing level of output x.

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Solution

Given cost function is : `C (x) = x^3/3 - 7x^2 + 111x + 50`

and demand function is : ` x = 100 - P ⇒ P = 100 -x`

(i) Revenue function,  `"R" (x) = Px = x (100 - x ) = 100 x - x^2`     ...(i)

(ii) Profit function,  `"P"(x)` = Revenue - Cost

                                          = R (x) - C (x)

= `100x  - x^2 - x^3/3 + 7x^2 - 111x - 50`

= `- x^3/3 + 6x^2 - 11x - 50`                                                      ...(ii)

(iii) Diffeerentiate equation (ii) w.r.t. x, we have

`(dP)/(dx)` =- x2 + 12x - 11                                                      .....(iii)

Now, `(dP)/dx `= 0 ⇒ -x2 + 12x - 11 = 0 ⇒ x2 - 12x + 11 = 0

⇒ (x - 1) (x - 11) = 0 ⇒ x = 1, 11
Again differentiate equation (iii), we have

`(d^2P)/(dx^2)` = 12 - 2x

At x = 1,  `(d^2P)/dx^2 = 10 ⇒ (d^2P)/dx^2 > 0   ("Minimum value")`

At x = 11,  `(d^2P)/dx^2 = -10 ⇒ (d^2P)/dx^2 < 0   ("Maximum value")`

Hence, the profit is maximum when output (x) is 11.

Concept: Application of Calculus in Commerce and Economics in the Cost Function
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