Final answer:
Critical values of the function y=102t+0.5t^2-t^3 are found by setting its first derivative, y' = 102 + t - 3t^2, equal to zero and solving for t within the interval [0,10].
Step-by-step explanation:
The critical values of a function occur where the first derivative is zero or undefined, indicating a local maximum, minimum, or point of inflection. For the function given as y=102t+0.5t2−t3, where 0≤t≤10, we first need to find its derivative, y' = 102 + t - 3t2. Setting the derivative to zero gives us the equation to solve:
0 = 102 + t - 3t2
By finding the roots of this quadratic equation, we will determine the critical values within the interval [0,10].