Elasticity of Demand Assignment help




Price Elasticity of Demand (PED or Ed) is a measure used in economics to show the responsiveness, or elasticity, of the quantity demanded of a good or service to a change in its price. More precisely, it gives the percentage change in quantity demanded in response to a one percent change in price (holding constant all the other determinants of demand, such as income). It was devised by Alfred Marshall. Price elasticities are almost always negative, although analysts tend to ignore the sign even though this can lead to ambiguity. Only goods which do not conform to the law of demand, such as Veblen and Giffen goods, have a positive PED. In general, the demand for a good is said to be inelastic (or relatively inelastic) when the PED is less than one (in absolute value): that is, changes in price have a relatively small effect on the quantity of the good demanded. The demand for a good is said to be elastic (or relatively elastic) when its PED is greater than one (in absolute value): that is, changes in price have a relatively large effect on the quantity of a good demanded. Revenue is maximised when price is set so that the PED is exactly one. The PED of a good can also be used to predict the incidence (or "burden") of a tax on that good. Various research methods are used to determine price elasticity, including test markets, analysis of historical sales data and conjoint analysis.

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PED is a measure of responsiveness of the quantity of a good or service demanded to changes in its price.The formula for the coefficient of price elasticity of demand for a good is:

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Application of Differentiation in Economics:-->

Economics deals with  cost and revenue functions.
The cost functions are average cost (AC), marginal cost (MC) and marginal average cost.
The revenue functions are  total revenue R, average revenue (AR) and marginal revenue (MR)
Profit is Revenue - cost.
So every economist will try to minimize the cost and maximise the revenue to get a maximum profit.
The cost functions has two components.  They are the fixed cost and the variable  cost.
Cost(C) is a function of output x . Hence C = f(x)
Average cost (AC) = total cost / output = C/x
Marginal cost is dC/dx
Marginal average cost is d(AC)/dx
Total revenue R = price x quantity sold  => R = px
Average revenue (AR) = total revenue / quantity = R/x
Marginal revenue (MR) =dR/dx = d(px)/ dx = p+ x* dp/dx

Application of Differentiation:- Elasticity of Demand

Differentiation  can be  applied to find the elasticity  of demand and supply.
Elasticity of demand:- If x = f(p) is the demand function, where x is demand and p is the price , then the elasticity of demand is  ηd = . dx  But demand is a decreasing function. Hence we introduce a negative sign and write ηd = -p . dx
                       x    dp                                                                                                                                    x   dp
Problem 1 :- A demand function is given by x = 25 - 4p - p2 . X is the demand for a good at price 'p'. Calculate the elasticity of demand.
Solution:-
Step 1 : Let us write the function x = 25 - 4p - p2
Step 2 : Let us now find dx/dp  =  0 - 4.1 - 2p =  -4 -2p
Step 3 : Let us now find ηd = - p . dx  =            -  p       . ( -4-2p) = >  4p + 2p2
                                             x   dp            25 - 4p - p2                         25 -4p -p2
Let us assume the price to be  3 then ηd4(3) + 2 (3)2 =   12 + 18  30  =  7.5
                                                               25- 4(3) - (3)2     25 -12-9      4
Hence elasticity of demand when p = 3 is 7.5

Find Elasticity of Supply

Let us apply differentitation to find the elasticity of supply.
Let 'x' be the supply and 'p' be the price , then x = f(p)
Elasticity of supply is given by the formula ηs = p . dx
                                                                    x    dp
#Problem 2 : Find the elasticity of supply for the supply function x = 2p2 + 5
Solution:-
Step 1 : Let us write the supply function x = 2p2 + 5
Step 2 : Let us differentiate the function dx = 2.2p + 0 = 4p
                                                          dp
Step 3 : Let us find ηs = p . dx  =       p       . 4p =          4p2    
                                                   2p2 + 5                2p2 + 5
If p = 3 then  ηs =  4 (3)2 / [2(3)2 + 5] =  36 / 23
Hence elasticity of supply when p =3 is  36/23 = 1.57

Maxmum Profit

Differentiation can be used in economics to maximize the profit.
# If Revenus  function is 2x2 + 10 x + 5 and cost function is 3x2 + 6x + 2 is for x units. Find the number of units to be sold at maximum profit.
Solution:-
Step 1 : Let us find the profit function which is Profit= Revenue  - cost = 2x2 + 10x + 5
                                                                                                    (-) 3x2  + 6x  + 2
                                                                                                              (-)      (-)
                                                                                                     _______________
                                                                                 Profit           =  -x2 + 4x  + 3
                                                                                                     _______________
Step 2 : Let us find dp/dx   =  - 2x + 4
Step 3 : Let us equate dp/dx = 0 =>  -2x = -4 => 2x = 4 => x = 2
Step 3 : Let us fnd d2p / dx= -2 < 0
            When d2p/dx2 < 0. the  profit is maximum. Hence profit is maximum when x = 2
Step 4: Substiting x  = 2 in the profit function we get the profit = - 22 + 4(2) + 3 = -4 +12 + 3 = 11
Answer : The number of units sold to get maximum profit is 2 and the profit is 11


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