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A company 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 help thanks.

Respuesta :

apologiabiology
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)=[tex] \frac{0-(2x-12)(125)}{(x^2-12x+61)^2}= \frac{1500-250x}{(x^2-12x+61)^2}[/tex]
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)=[tex] \frac{1}{ \sqrt{2x+1} } [/tex]
P'(x)=[tex] \frac{1500-250x}{(x^2-12x+61)^2}-\frac{1}{ \sqrt{2x+1} }[/tex]
find zeroes or what value of x make P'(x) equal to 0
[tex]\frac{1}{ \sqrt{2x+1} }= \frac{1500-250x}{(x^2-12x+61)^2}[/tex]
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