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.
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
