. A one-product firm estimates that its daily total cost function (in suitable units) is C (x) = x^3 - 6x^2 + 13x + 15 and its total revenue function is R(x) = 28x. Find the value of x that maximizes the daily profit. (4 points)

Value of x maximising profit : x = 5

Explanation:

Cost : C(x) = x^3 - 6x^2 + 13x + 15 ; Revenue: R(x) = 28x

Profit : Revenue - Cost = R(x) - C(x)

28x - [x^3 - 6x^2 + 13x + 15]  = 28x - x^3 + 6x^2 - 13x - 15

= - x^3 +  6x^2 + 15x - 15

To find value of 'x' that maximises total profit , we differentiate total profit function with respect to x & find that x value.

dTP/dx = - 3x^2 + 12x + 15 = 0 ► 3x^2 - 12x - 15 = 0

3x^2 + 3x - 15x - 15 = 0  ► 3x (x +1) - 15 (x + 1) = 0 ► (x+1) (3x-15) = 0

x + 1 = 0 ∴ x = -1   [Rejected, production quantity cant be negative] ;

3x - 15 = 0 ∴ 3x = 15 ∴ x = 15/3 = 5

Double derivate : d^2TP/dx^2 = - 6x + 12

d^2TP/dx^2  i.e - 6x + 12  at x = 5 is -6(5) + 12 = - 30+ 12 = -8 which is negative. So profit function is maximum at x = 5

the most important factor in the

work process or industry is to retain the interest of the employee. and to

retain them, their salary or profit must be adjusted to the best value. unless

the employee does not do his job properly, employees must receive bonuses or

benefits to ensure them to stay in the company. it will actually make them stay

longer, make them feel important and reduce costs for hiring new employees.

explanation:

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