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Fofino [41]
3 years ago
6

Revenue from Monorail Service, Las Vegas In 2005 the Las Vegas monorail charged $3 per ride and had an average ridership of abou

t 28,000 per day. In December 2005 the Las Vegas Monorail Company raised the fare to $5 per ride, and average ridership in 2006 plunged to around 19,000 per day.
a. Use the given information to find a linear demand equation.
b. Find the price the company should have charged to maximize revenue from ridership. What is the corresponding daily revenue?
c. The Las Vegas Monorail Company would have needed $44.9 million in revenues from ridership to break even in 2006. Would it have been possible to break even in 2006 by charging a suitable price?
Mathematics
1 answer:
Karolina [17]3 years ago
5 0

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 = -\frac{b}{2a}  = - \frac{41500}{2(-4500)} =  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

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The plane will end up flying 5.02°.

The plan's speed relative to the ground will be 645.91 km/hr.

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Use the cosine formula,

a^{2}=b^{2}+c^{2}-2 b c \cos A

Substitute the given values in the formula,

\mathrm{R}^{2}=700^{2}+80^{2}-(2 \times 700 \times 80 \times \cos 45^\circ)

\mathrm{R}^{2}=490000+6400-(112000 \times\frac{1}{\sqrt{2} } )

\mathrm{R}^{2}=496400-79195.95

\mathrm{R}^{2}=417204.049

Taking square root on both sides, we get

R = 645.91 km/hr

This is the grouped speed of the aircraft.

To find θ use sine rule.

$\frac{\sin C}{c}=\frac{\sin A}{a}

$\frac{\sin \theta}{80}=\frac{\sin 45}{645.91}

Do cross multiplication, we get

${\sin \theta}}=\frac{\sin 45}{645.91}\times 80

${\sin \theta}}=\frac{\frac{1}{\sqrt{2} } }{645.91}\times 80

sin θ = 0.0875

θ = 5.02°

This is known as the drift angle and is the correction the pilot should apply to remain on course.  

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The track is the actual direction over the ground which is θ = 5.02°

An alternative method to this would be to separate each vector into vertical and horizontal components and add.

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