I don't know how to do it except for with deritivies
so take the deritivieve and find where the deritivieve equals 0
that is where the sign changes
where the sign changes from (+) to (-), that is max
so
A.
max revenue
R'(x)=
![\frac{0-(2x-12)(125)}{(x^2-12x+61)^2}= \frac{1500-250x}{(x^2-12x+61)^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B0-%282x-12%29%28125%29%7D%7B%28x%5E2-12x%2B61%29%5E2%7D%3D%20%5Cfrac%7B1500-250x%7D%7B%28x%5E2-12x%2B61%29%5E2%7D)
find where numerator is 0
at x=6
to find change of sign, evaluate the denomenator at above and below 6 and see sign
R'(5)=(+)
R'(7)=(-)
at x=6, the sign changes from (+) to (-)
max is at x=6
sub 6 for x in the R(x) function
R(6)=9 (it's in thousands so $9000 is te max revenue)
B.
max profit
combine them
P(x)=R(x)-E(x)
take the deritive of P(x)
using sum rule
P'(x)=R'(x)-E'(x)
we already know what R'(x) is
E'(x)=
![\frac{1}{ \sqrt{2x+1} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2x%2B1%7D%20%7D%20)
P'(x)=
![\frac{1500-250x}{(x^2-12x+61)^2}-\frac{1}{ \sqrt{2x+1} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B1500-250x%7D%7B%28x%5E2-12x%2B61%29%5E2%7D-%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2x%2B1%7D%20%7D)
find zeroes or what value of x make P'(x) equal to 0
![\frac{1}{ \sqrt{2x+1} }= \frac{1500-250x}{(x^2-12x+61)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2x%2B1%7D%20%7D%3D%20%5Cfrac%7B1500-250x%7D%7B%28x%5E2-12x%2B61%29%5E2%7D)
use calculator or something or work it out to find x
at x=5.225
x is hundreds so times 100
522.5
about 523 items
A. $9000
B. 523 items