A model for a company's revenue from selling a software package is R(p) = -2.5p² + 500p, where p is the price in dollars of the software. What price will maximize revenue? Find the maximum revenue.

The price $_____ maximizes the revenue.
The maximum revenue is $_____.

Respuesta :

Answer: p = $100,  R = $25,000

Step-by-step explanation:

The maximum is the y-value of the Vertex.

Step 1: Use the Axis-Of-Symmetry (AOS) formula to find x: [tex]x=\dfrac{-b}{2a}[/tex]

R(p) = -2.5p² + 500p

        a=-2.5  b=500

[tex]p=\dfrac{-(500)}{2(-2.5)}[/tex]

   [tex]=\dfrac{-500}{-5}[/tex]

   [tex]=100[/tex]


Step 2: Find the maximum by plugging the p-value (above) into the given equation.

R(100) = -2.5(100)² + 500(100)

          = -25,000 + 50,000

          = 25,000

gmany

It's a quadratic function

[tex]f(x)=ax^2+bx+c[/tex]

If a < 0, then the function has a maximum value in vertex.

The vertex (h, k), where:

[tex]h=\dfrac{-b}{2a},\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[/tex]

We have:

[tex]R(p)=-2.5p^2+500p\\\\a=-2.5<0,\ b=500,\ c=0[/tex]

Substitute:

[tex]h=\dfrac{-500}{2(-2.5)}=\dfrac{-500}{-5}=100\\\\k=R(100)=-2.5(100)^2+500(100)\\\\k=-2.5(10,000)+50,000\\\\k=-25,000+50,000\\\\k=25,000[/tex]

Answer:

The price $100 maximizes the revenue.

The maximum revenue is $25,000.