Suppose a monoploy with total cost function CQ 1 2Q Q2 Th

Suppose a monoploy with total cost function C(Q) = 1 + 2Q + Q^2.

The linear inverse demand function for its product is given by P = 8 - 2Q.

(a) Find the fixed cost of the monopoly.

Fixed Cost = 1

(b) Write down the total revenue function R(Q) and the profit function (Q)

Total revenue, R(Q) = P x Q = 8Q - 2Q²

Profit, (Q) = R(Q) - C(Q) = 6Q - 3Q² - 1

(c) Write the profit function in verterx form, compute the maximum profit and the associated quantity and price.

I need help to do question (c). please do it.

Solution

the profit function is :

(Q) = R(Q) - C(Q) = 6Q - 3Q² - 1

the profit function is of degree 2 that is its parabolic

-3Q^2 + 6Q - 1 is the standard for of the profit fucntion

the profit function is of the form:

y = ax^2 + bx + c

where a = -3 , b = 6 and c = -1

the vertex form is given as

y = a(x-h)^2 + k

where h and k are the x and the y coordinates of the vertes of the parabola

so first we need to fins the vertex of the profit function

the x coordinate of the vertex is given as x = -b/2a = -6/(2*-3) = 1

now for the y coordinate of the vertex plug x = 1 in the profit function

6Q - 3Q² - 1         -------{Q is same as x}

(Q) = 6(1) - 3(1)^2 -1 = 2

hecen h = 1 and k = 2

=> y = -3(x - 1) + 2

replace x by Q and y by (Q)

=> (Q) = -3(Q - 1) + 2   ---------> this is the vertex form of the profit function

as the profit fucntion has a parabolic curve so the maximum profit would be attained at the vertex (h,k)

therefore (Q) = 2 is the maximum profit and the quantity associated with it is Q = 1