find the minimum value of ( f(t)=3t^{3}-27t^{2}+40), (tgeq0), and give the value of (t) where this minimum…

Explanation:

Step1: Find the derivative

Differentiate $f(t)=3t^{3}-27t^{2}+40$ using the power - rule. The derivative $f^\prime(t)=9t^{2}-54t = 9t(t - 6)$.

Step2: Find the critical points

Set $f^\prime(t)=0$. So, $9t(t - 6)=0$. Solving this gives $t = 0$ and $t=6$ as critical points.

Step3: Use the second - derivative test

Differentiate $f^\prime(t)$ to get the second - derivative $f^{\prime\prime}(t)=18t-54$. For $t = 0$, $f^{\prime\prime}(0)=- 54<0$, so $t = 0$ is a local maximum. For $t = 6$, $f^{\prime\prime}(6)=18\times6 - 54=54>0$, so $t = 6$ is a local minimum.

Step4: Find the minimum value

Substitute $t = 6$ into the original function $f(t)$. $f(6)=3\times6^{3}-27\times6^{2}+40=3\times216-27\times36 + 40=648-972 + 40=-284$.

Answer:

The minimum value occurs when $t = 6$ with a value of $-284$.