Answer:
maximum at 7.50 rate with 484
Step-by-step explanation:
Given that a babysitting club sits for 50 different families. They would like to
increase their current rate of $9.50 per hour.
As of now revenue =[tex]50(9.50) = 475[/tex]
Also given that after surveying the families, the club finds that the number of families will decrease by about 2 for each $0.50 increase in the hourly rate.
Let x be the no of times 0.50 increase is affected. Then no of families would be 50-2x
Hence revenue as a function of x is
f(x) =[tex](9.50+0.50x)(50-2x)\\= 475 +6x-x^2[/tex]
x can take values between 0 and 25.
Range can be (0,484)
(vertex of parabola of f(x) since open down)
c) [tex]f'(x) = 6-2x\\f"(x) =-2<0[/tex]
Equate I derivative to 0
x =3
Hourly rate at [tex]9-3(0.50)\\=7.50[/tex]
will maximize the revenue.
Yes. reasonable
because within the domain
d) Maximum income is 484