optimization of shaded area with interval

Question

Find the x-value maximizing the shaded area on the interval 0 <= x <= 22. One vertex is on the graph of f(x) = x^2/3 - 50x + 1000.

x = ?

Answer 1

As I understand you graph the function in the interval [0,22] and see where it reaches its maximum value.

Function f(x) =x^2/3-50x+1000 reaches max(min) values either on extreme (critical) points or endpoints. You can take derivative and find critical point(s), look hoe function behaves on [0, 22] (increases, decreases, etc.) Then compute values on maximum and endpoints.

A critical point of f(x) by derivative would be x=75, which outside of the interval [0, 22]. So now we need to see f(x) is increasing or decreasing on [0,22]. If it is increasing, then the max will be at x=22.

Give a try.