Max spends three times as long on his math homework than he does on his science homework. If he spent total of 64 minutes on mat
2 answers:
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Answer:
See explanation
Step-by-step explanation:
Solution:-
- The demand function of a certain brand is given as price P a function of x quantity of goods ( in hundred ) demanded per month. The relation is:
P ( x ) = 50 Ln ( 50 / x ).
- The point price elasticity ( E ) of demand is given by:
- Where, dP / dx : is the rate of change of price ( P ) with each hundred unit of good ( x ) is demanded.
- To determine the " dP / dx " by taking the first derivative of the given relation:
P ( x ) = 50 Ln ( 50 / x ).
d P ( x ) / dx = [ 50*x / 50 ] * [ -1*50 / x^2 ]
= - 50 / x
- Hence the point price elasticity of demand is given by:
E = - ( P / x ) * ( 50 / x )
E = -50*P / x^2
- For an inelastic demand, ! E ! is < 1:
! -50*P / x^2 ! < 1
50*P / x^2 < 1
P < x^2 / 50
- For an unitary demand, ! E ! is = 1:
! -50*P / x^2 ! = 1
50*P / x^2 = 1
P = x^2 / 50
- For an inelastic demand, ! E ! is > 1:
! -50*P / x^2 ! > 1
50*P / x^2 > 1
P > x^2 / 50
2)
If the unit price is increased slightly from $50, will the revenue increase or decrease?
- We see from the calculated demand sensitivity d P / dt:
d P ( x ) / dx = - 50 / x
- We see that as P increases the from P = $50, the quantity of goods demanded would be:
50 = 50 ln(50/x)
1 = Ln ( 50 / x )
50/x = e
x = 50 / e
Then,
d P ( x ) / dx = - 50 / ( 50 / e )
d P ( x ) / dx = - e
- We see that if price slightly increases from $ 50 then the quantity demanded would decrease by e (hundreds ) goods.
- The decrease in the quantity demanded is higher than the increase in price. The revenue is given by the product of price P ( x ) and x:
Revenue R ( x ) = P ( x ) * x
= 50*x*ln(50/x)
Then the product of price and quantity goods also decreases; hence, revenue decreases.