Find the minimum and maximum values of the function $y=12 t^{3}+18 t^{2}$ on $[1,2]$ by comparing values at the critical points and endpoints. Submit your answers as exact values. Do not round. \[ \begin{array}{l} y_{\min }= \\ y_{\max }= \end{array} \]
Solution
Step 1 ：First, we need to find the derivative of the function \(y = 12t^3 + 18t^2\).
Step 2 ：The derivative of the function is \(y' = 36t^2 + 36t\).
Step 3 ：Next, we find the critical points of the function by setting the derivative equal to zero and solving for \(t\). The critical points are \(t = -1\) and \(t = 0\).
Step 4 ：However, since the interval we are considering is \([1,2]\), we only need to evaluate the function at the endpoints of this interval and at any critical points that fall within this interval. In this case, there are no critical points within the interval, so we only need to evaluate the function at \(t = 1\) and \(t = 2\).
Step 5 ：Evaluating the function at these points, we find that \(y(1) = 30\) and \(y(2) = 168\).
Step 6 ：Thus, the minimum value of the function on the interval \([1,2]\) is \(\boxed{30}\) and the maximum value is \(\boxed{168}\).
