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Basic Mathematics – I
Notes
1
Thus p 10 x 25
3.75
or x – 25 = – 3.75(p – 10) or x = 62.5 – 3.75p is the required equation.
2
(ii) The given demand function is p = 3x – 100x + 800
dp
= 6x – 100 = 60 – 100 = –40 when x = 10
dx
When x = 10, we have p = 300 1,000 + 800 = 100
dx p 1 100 1
=
d
dp x 40 10 4
When price increases by 4%, then the approximate change in demand in given by the formula
% change in demand
=
d % change in price
or % change in demand = – × % change in price
d
1
= 0.04 = –0.01
4
i.e. demand falls by 1%.
The new demand = old demand × 0.99
= 10 × .99 = 9.9
New price = 100 × 1.04 = 104
dx p 1 p
= .
d
dp x (6x 100) x
1 104
= 0.2587
(6 9.9 100) 9.9
Example
2
(i) If x = 2Y , find income-elasticity of demand.
p B 1
(ii) If x A , find cross-elasticity of demand when p = 5.
p 2 B
B
Solution:
dx Y Y
(i) = 4Y 2
Y 2
dY x 2Y
dx p
(ii) = A B
AB
dp x
B A
dx A p B 2 p B 1 3 1
Now = at p = 5
dp 2 2 3 B
B p B 2 p B 2
1 5 5
Also x = 2 when p = 5 = –0.83.
A B AB
3 2 6
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