Answer to Question #131962 in Economics of Enterprise for Rnlf

Question

A company has established that the relationship between sale price for one of its product and the quantity sold per month is approximately D = 780 -10p units (D is the demand or quantity sold per month, and p is the price in pesos). The fixed cost is P800 per month, and the variable cost is P30 per unit produced. What number of units, D*, should be produced per month and sold to maximize net profit? What is the maximum profit per month related to the product?

Answer 1

ANSWERS

D* = 240 units per month

Maximum profit (π) = P4,960 per month

SOLUTIONS

We are given $D = 780 - 10P \space units$

$Fixed \space costs (FC) = P800 \space per \space month$

$Variable \space costs = P30 \space per \space unit$

The inverse monthly demand function is therefore given by:

$10P = 780 - D$

$P = \dfrac {780}{10} - \dfrac {D}{10}$

$P = 78 - 0.10D$

$Price (P) = Average \space revenue (AR)$

$\therefore AR = 78 - 0.10D$

Total revenue (TR) is given by:

$TR = AR × D$

$= (78 - 0.10D) × D$

$= 78D - 0.10D^2$

Total cost = Variable costs + Fixed costs

$TC = VC + FC$

$= (P30 × D) + P800$

$= 800 + 30D$

Total profit is given by:

Total profit = Total revenue - Total costs

$π = TR - TC$ $π = (78D - 0.10D^2) - (30D + 800)$

$= 78D - 30D - 0.10D^2 - 800$

$= -800 + 48D - 0.10D^2$

Maximum profit is approached through differential calculus

$\dfrac {dπ}{dD} = \dfrac {d}{dD}(-800 + 48D - 0.10D^2)$

$= 48 - 0.20D$

$\dfrac {d^2π}{dD^2}= \dfrac {d^2}{dD^2}(48-0.20D)$

$= -0.20$

$Since \space \dfrac {d^2π}{dD^2} < 0$ , profit(π) is maximum.

When profit (π) is maximum,

$\dfrac {dπ}{dD} = 0$

$=> 48 - 0.20D = 0$

$0.20D = 48$

$D = \dfrac {48}{0.20}$

$= 240 \space units$

$\therefore D^* = 240 \space units \space per \space month$

Calculating maximum profit (π)

$Max(π) = -800 + 48D^* - 0.10(D^*)^2$

$= -800 + 48(240) - 0.10(240^2)$

$= -800 + 11 520 - 5760$

$= P4,960$

$\therefore Max(π) = P4,960 \space per \space month$