Revenue function question.?

Question

Selling 60x - x^2-500 trees a day I need to find maximum revenue.

A.) I need a revenue function where x = price per tree
B.) I need a maximum and minimum price to sell any trees
C.) What price maximizes revenue?

Thanks,

Answer 1

A.) Your revenue R would be R = PT, where P is price per tree and T is number of trees sold. You were given a function that is equivalent to T = 60P - P^2 - 500, so R = PT = x(60x - x^2 - 500) = -x^3 + 60x^2 - 500x.

B.) The minimum and maximum price to sell any trees would be the x-values where T = 0. 60x - 500 - x^2 = 0 ==> x^2 - 60x + 500 = 0 ==> (x - 50)(x - 10) = 0 ==> x = 50 or x = 10, so the minimum price is 10 and the maximum price is 50, although it's not clear why people would refuse to buy trees for less than $10.

C.) R = -x^3 + 60x^2 - 500x, and any continuous function has extrema when its derivative is equal to zero. dR/dx = -3x^2 + 120x - 500 = 0 ==> x = (-120 +/- sqrt(120^2 - 4*(-3)*(-500))) / 2*(-3) = (-120 +/- sqrt(14400 - 6000)) / -6 = (-120 +/- 91.65) / -6 = 4.72 or 35.3. Since 4.72 falls below the cutoff of 10 we found in B.), the answer must be $35.3 / tree.