y = Ax ( 3L^2 − 4x^2 )

Question

L=10 and A=8

1. Evaluate dy/dx
2. Where is dy/dx = 0, in other word, for which x does the maximum deflection occur, x=?
3. Find the maximum delfelction, y=?

Answer 1

Plugging the values A=8 and L=10 we get the function y(x)=2400x-32x^3 which implies that the derivative of y(x) equals y'(x)=2400-96x^2. In order to find the maximum we need to solve the inequality y'(x)>=0 which is the same as 96x^2<=2400 or x^2<=25 or |x|<=5. Therefore, the derivative of y(x) is negative on the left of -5, positive between -5 and 5 and negative again on the right of 5. This means that there is a local maximum of y(x) at the value x=5 which gives y(5)=12000-4000=8000.