Acompany that produces electronic components can model its revenue and expense by the functions r(x)= (125/(x^2-12x+61))+4 and e(x) = sqrt(2x+1)+3 respectively, where x is hundreds of components produced and r(x) and e(x) are in thousands of dollars. assuming 0 ≤ x ≤ 10, answer the following.

a) to the nearest dollar, what is the maximum revenue?
b) if profit is calculated as the difference between revenue and expense, p(x) = r(x) - e(x), how many items should be produced to maximize profit?

i missed class the day we went over min and max and am very confused on this problem and would really appreciate .

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)=
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)=
P'(x)=
find zeroes or what value of x make P'(x) equal to 0

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

50.2450.24 in squared

Answer:

$4034.40

step-by-step explanation:

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40