A monopolist has a demand curve x = 106 – 2p and average cost curve AC = 5 + x/50, where p is the price per unit output

Question

A monopolist has a demand curve x = 106 – 2p and average cost curve AC = 5 + \(\frac{x}{50}\), where p is the price per unit output and x is the number of units of output. If the total revenue is R = px, determine the most profitable output and the maximum profit.

Answer 1

x = 106 – 2p 

(or) 2p = 106 – x 

p = \(\frac{1}{2}\)(106 – x) 

Revenue, R = px 

= \(\frac{1}{2}\)(106 – x) x 

= 53x – \(\frac{x^2}{2}\)

Average Cost, AC = 5 + \(\frac{x}{50}\)

Cost C = (AC)x

= (5 + \(\frac{x}{50}\))x

= 5x + \(\frac{x^2}{50}\)

Profit (P) = Revenue – Cost 

\(\frac{dP}{dx}\) = 48 - \(\frac{13(2x)}{25}\)

\(\frac{dP}{dx}\) = 0 gives 

48 – \(\frac{13(2x)}{25}\) = 0 

48 = \(\frac{13\times2x}{25}\)

x = \(\frac{13\times25}{13\times2}\) = 46.1538 = 46 (approximately) 

Also \(\frac{d^2P}{dx^2}\) = 0 - \(\frac{(13)^2}{25}\), negative since \(\frac{d^2P}{dx^2}\) is negative, 

Profit is maximum at x = 46 units. 

Profit = 48x – \(\frac{13}{25}\)x² 

When x = 46, 

Profit = 48 x 46 – \(\frac{13}{25}\) x 46 x 46 

= 2208 – \(\frac{27508}{25}\)

= 2208 – 1100.32 

= Rs 1107.68