Maximums and Revenue

Question

The demand function for a product is p=32-2q where p is the price in dollars when q units are demanded. Find the level of production that maximizes the total revenue and determine the maximum revenue. q= ___ units & R= $_____ 

Answer 1

Revenue = (number of units sold)·(price per unit)

 

R = q·(32-2q)

R = 32q - 2q²

 

This can be solved algebraically and with Calculus. 

 

1)  Algebraically, the revenue function is a quadratic equation whose graph is an inverted parabola.  The vertex of the parabola with thus be the maximum point on the Revenue graph.  Find the vertex by putting the quadratic into the vertex form, y = a(x-h)² + k, where (h,k) is the location of the vertex.  To convert to vertex form, you will need to "complete the square" on the revenue function.

 

R = -2q² + 32q

R = -2(q² - 16q + (16/2)²) + 2(16/2)²

R = -2(q² - 16q + 64) + 128

R = -2(q-8)² + 128

 

The vertex is located at (8, 128).  The value of q that maximizes revenue is q = 8 and the max revenue is R = 128

 

2) With calculus, take the derivative of R wrt q, set it to zero, and solve for q:

 

R = -2q² + 32q

dR/dq = -4q + 32

0 = -4q + 32

4q = 32

q = 8    (Units sold which gives max revenue)

 

R(8) = 32(8) - 2(8)²

R(8) = 256 - 128 = 128   (Max revenue)