Question

An online music store sells about 4000 songs each day when it charges $1 per song. For each $0.05 increase in price, about 80 fewer songs per day are sold. Use the verbal model and quadratic function to determine how much the store should charge per song to maximize daily revenue. R(x)=(1+0.05x)(4000-80x)

Answer 1

Answer:

The best price is $1.75.

Step-by-step explanation:

Profit = Songs x Price  

Given equation is ;

$R(x)=(1+0.05x)(4000-80x)$

Solving this :

Let the number of $0.05 increases be x.

R(x) = $-4x^{2}+120x+4000$

The revenue is maximized when dR/dx = 0.  

dR/dx = $-8x+120$

The revenue is maximized when x = 15

Means when the price per song = $1.00 + ($0.05*15) = $1.75

Therefore, the best price is $1.75.

Answer 2

The store should charge $1.75 to maximize daily profit.

The function which is required to be maximised is;

• R(x)=(1+0.05x)(4000-80x)

• R(x)=(1+0.05x)(4000-80x)R(x) = -4x² + 120x + 4000

The function is maximized when it's gradient (dR/dx) = 0

• dR/dx = -8x + 120

• 0 = -8x + 120

• 8x = 120

• x = 15

Therefore, the best price is at;

= (1+0.05×15)

= $1.75

Therefore, the store should charge $1.75 per song to maximize daily revenue.

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