Question 210960
{{{R = 1500x-3x^2}}} where {{{x<25}}} Substitute R = 17,568 and solve for x.
{{{17568 = 1500x-3x^2}}} Rearrange into standard quadratic form.
{{{3x^2-1500x+17568 = 0}}} Divide through by 3 to ease the calculations a bit.
{{{x^2-500x+5856 = 0}}} Factor.
{{{(x-488)(x-12) = 0}}} so...
{{{x = 488}}} or {{{x = 12}}} but {{{x < 25}}}, so...
{{{highlight(x = 12)}}} is the answer.