Solve y=10x^3 + 5.5x^2 - 58.50x + 45.00
2 answers:
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Answer:
A)The required linear demand equation ( q ) = -4500p + 41500
B) $4.61
$95680.55
C) No it would not have been possible by charging a suitable price
Step-by-step explanation:
<u>A) find the linear demand equation</u>
given two points ; ( 3, 28000 ) and ( 5, 19000 )
slope ( m ) = ( y2 - y1 ) / ( x2 - x1 )
= ( 19000 - 28000 ) / ( 5 - 3 ) = -4500
slope intercept is represented as ; y = mx + b
where y( 28000) = -4500(3) + b
hence b = 41500
hence ; y = -4500x + 41500
The required linear demand equation ( q ) = -4500p + 41500 ----- ( 1 )
p = price per ride
<u>B ) Determine the price the company should charge to maximize revenue from ridership and corresponding daily revenue</u>
Total revenue ( R ) = qp
= p ( -4500p + 41500 )
hence R = -4500p^2 + 41500p ------ ( 2 )
To determine the price that should maximize revenue from ridership we will equate R = -4500p^2 + 41500p to a quadratic equation R(p) = ap^2 + bp + c
where a = -4500 , b = 41500 , c = 0
hence p = =
= 4.61
$4.61 is the price the company should charge to maximize revenue from ridership
corresponding daily revenue = R = -4500p^2 + 41500 p
where p = $4.61
hence R = -4500(4.61 )^2 + 41500(4.61) = $95680.55
C) No it would not have been possible by charging a suitable price