Respuesta :
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:
[tex]R(x)=(5.50-0.50x)(320+40x)[/tex]
To find:
The maximum potential revenue.
Solution:
We have,
[tex]R(x)=(5.50-0.50x)(320+40x)[/tex]
It can be written as:
[tex]R(x)=(5.50)(320)+(5.50)(40x)+(-0.50x)(320)+(-0.50x)(40x)[/tex]
[tex]R(x)=1760+220x-160x-20x^2[/tex]
[tex]R(x)=1760+60x-20x^2[/tex]
Differentiate with respect to x.
[tex]R'(x)=0+60(1)-20(2x)[/tex]
[tex]R'(x)=60-40x[/tex]
For critical points, [tex]R'(x)=0[/tex].
[tex]60-40x=0[/tex]
[tex]60=40x[/tex]
[tex]\dfrac{60}{40}=x[/tex]
[tex]1.5=x[/tex]
Differentiate R'(x) with respect to x.
[tex]R''(x)=0-40(1)[/tex]
[tex]R''(x)=-40[/tex]
Since R''(x)<0, therefore function R(x) is maximum at [tex]x=1.5[/tex]. The maximum value is:
[tex]R(1.5)=1760+60(1.5)-20(1.5)^2[/tex]
[tex]R(1.5)=1760+90-45[/tex]
[tex]R(1.5)=1805[/tex]
Therefore, the maximize potential revenue is $1805 at [tex]x=1.5[/tex].
