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Answer:
a) 0.3 ( Inelastic )
b) 1.5 ( elastic )
c) $0.66 = ( 1/1.5 )
d) $44342.77
e) 66514.1633 Ib
Explanation:
Function = q (p) = 100e^1.5 ( 5 − p )
∴ q = 100e^(7.5 - 1.5p)
dq/dp = 100 ( -1.5 ) e^(7.5-1.5p)
hence E (p ) = | p/q * dq/dp |
= | -1.5p |
a) Determine the price elasticity of demand at $0.20 / Ib
E( 0.2 ) = | -1.5 * 0.2 |
= 0.3
given that E < 1 inelastic more revenue is generated when prices are increased
b) price elasticity at $1 / Ib
E ( 1 ) = | -1.5 * 1 |
= 1.5 given that E > 1 ( Elastic ) , hence when prices are raised less revenue is generated
c) Determine price at which maximum revenue is achieved
i.e. E(P) = -1 ( for max revenue )
-1 = | -1.5p |
therefore p ( price at which max revenue is achieved ) = ( 1 / 1.5 ) = $0.66
d) maximum revenue per week
q * p = ( 0.66 )* 100*e^(7.5 - 1.5*0.66)
= $44342.77
e) How many pounds will be sold each week at optimal price
quantity sold at optimal price ( q )
q = 100*e^(7.5 - 1.5*0.66)
= 66514.1633