Question

Lavro operates a popular hotdog stand. He sells 320 sausages per day at a price of

$5.50 each. The previous year's sales show that for every $0.50 decrease in price, he
will sell another 40 sausages.
The equation that models this problem is:
R(x) = (price)
(number sold)
R(x) = (5.50 - 0.50x)(320 + 40x)
Question 3 (1 point)
What is his maximize potential revenue?

Answer 1

Given:

Lavro sells 320 sausages per day at a price of $5.50 each.

The previous year's sales show that for every $0.50 decrease in price, he will sell another 40 sausages.

The equation that models this problem is:

`R(x)=(5.50-0.50x)(320+40x)`

To find:

The maximum potential revenue.

Solution:

We have,

`R(x)=(5.50-0.50x)(320+40x)`

It can be written as:

`R(x)=(5.50)(320)+(5.50)(40x)+(-0.50x)(320)+(-0.50x)(40x)`

`R(x)=1760+220x-160x-20x^2`

`R(x)=1760+60x-20x^2`

Differentiate with respect to x.

`R'(x)=0+60(1)-20(2x)`

`R'(x)=60-40x`

For critical points,
`R'(x)=0`.

`60-40x=0`

`60=40x`

`(60)/(40)=x`

`1.5=x`

Differentiate R'(x) with respect to x.

`R''(x)=0-40(1)`

`R''(x)=-40`

Since R''(x)<0, therefore function R(x) is maximum at
`x=1.5`. The maximum value is:

`R(1.5)=1760+60(1.5)-20(1.5)^2`

`R(1.5)=1760+90-45`

`R(1.5)=1805`

Therefore, the maximize potential revenue is $1805 at
`x=1.5`.