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11111nata11111 [884]
3 years ago
5

Pls help. Will mark brainliest

Mathematics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

see below (I hope this helps!)

Step-by-step explanation:

p² = 4x² * (q(1 + r² / y²) / S)

p²S = 4x² * q(1 + r² / y²)

p²S / 4x²q = 1 + (r² / y²)

p²S / 4x²q - 1 = r² / y²

y² = r² / (p²S / 4x²q - 1)

y = √( r² / (p²S / 4x²q - 1) = √(r²4x²q / (p²S - 4x²q))

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A)The required linear demand equation ( q ) = -4500p + 41500

B) $4.61

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Step-by-step explanation:

<u>A)  find the linear demand equation</u>

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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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Answer:

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