Find the point of diminishing returns for the function, where R is the revenue (in thousands of dollars) and x is the advertising cost (in thousands of dollars).

Question

R( x )= 1/50,000(600x^2-x^3)

Answer 1

max Revenue is where

R'(x) =0

1/(50(.6x^2 -x^3)

=(30x^2 -50x^3)^-1

derivative = -(30x^2-50x^3)(60x-150x^2)/(30x^2-50x^3)^2

= -(60x-150x^2)/(30x^2-50x^3)

=-(6-15x)/(3x -5x^2) =0

6-15x = 0

x = 6/15 = 2/5 = 0.4 thousand dollars = $400

> $400 is the point of diminishing revenue returns

or

R(x) =(1/50)(.6x^2 -x^3)

R'(x) = .02(.12x -3x^2)

= .024x - .06x^2 = 0

x(.024 - .06x) = 0

x = .024/.06 = 24/60 = 4/10 =.4 thousand dollars = $400 as the point of diminishing returns

$400 either way

don't spend more than $400 on advertising to max out Revenue

BUT that's max revenue, not maximum profit

last twist on the problem

P(x) = R(x) -C(x)

find where

P'(x) = R'(x)-C'(x) = 0 = MR= MC = Marginal Revenue = Marginal Cost

=.024x -.06x^2= x' = 1

.06x^2 -.024x =1

x^2 -.4x =1/.06 = 16 2/3

x^2 -.4x + .2^2 = -16 2/3 +.04

(x-.2)^2 = 17.0666...

x = 2 +/-sqr(17.0666...)

x = about 2+4.13118235

x = about 6.13 = $6,131,18 is the point of diminishing returns for advertising expenditures

(no guarantees on the arithmetic above. but it's the general method for finding the solution. MR=MC, any output is the point of diminishing returns of profit)