Question

ANSWER A AND B a. assuming a linear demand curve find the price of hamburger that maximize the nightly hamburger revenueB. if the concession had to fix cost of $2500 per night in the variable cost of $.60 per hamburger find the price of a hamburger that will maximize a nightly hamburger profit

Answer 1

A.

When the price was 6.10, the number of hamburgers sold was 4750.

When the price was 6.70, the number of hamburgers sold was 3250.

That means an increase of 0.6 in price caused the number of hamburgers sold to decrease by 1500.

If the price increases x times 0.6, the number of hamburgers sold will decrease x times 1500.

Since the revenue y is given by the price times the number of units sold, we have:

`\begin{gathered} y=(6.10+0.6x)(4750-1500x)\\ \\ y=28975-9150x+2850x-900x^2\\ \\ y=-900x^2-6300x+28975 \end{gathered}`

The maximum value is given by the vertex of the parabola:

`x_v=(-b)/(2a)=(6300)/(-1800)=-3.5`

Therefore the price of a hamburger that maximizes the revenue is:

`6.1+0.6(-3.5)=6.1-2.1=4`

The price is $4.00.

B.

To find the new best price, let's calculate the profit including the new costs:

`\begin{gathered} y=-900x^2-6300x+28975-2500-0.6(4750-1500x)\\ \\ y=-900x^2-6300x+26475-2850+900x\\ \\ y=-900x^2-5400x+23625\\ \end{gathered}`

The new x-coordinate of the vertex is:

`x_v=(-b)/(2a)=(5400)/(-1800)=-3`

And the new best price is:

`6.1+0.6(-3)=6.1-1.8=4.3`

The new price is $4.30.