If `S=(t^(3))/(3)-2t^(2)+3t+4`, then

Options

A. at t = 1 , S is minimum
B. at t = 1 , S is maximum
C. at t = 3 , S is maximum
D. at t = 2 , S is minimum

Answer

To find the maximum and minimum points of the function \( S(t) = \frac{t^3}{3} - 2t^2 + 3t + 4 \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( S(t) \) with respect to \( t \):

\[ \frac{dS}{dt} = \frac{d}{dt}\left(\frac{t^3}{3} - 2t^2 + 3t + 4\right) \]

Using the power rule of differentiation, we compute:

\[ \frac{dS}{dt} = t^2 - 4t + 3 \]

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find the critical points:

\[ t^2 - 4t + 3 = 0 \]

### Step 3: Solve the quadratic equation
We can solve this quadratic equation using the factorization method:

\[ (t - 1)(t - 3) = 0 \]

Thus, the solutions are:

\[ t = 1 \quad \text{and} \quad t = 3 \]

### Step 4: Find the second derivative
Now, we need to determine whether these critical points correspond to a maximum or minimum by computing the second derivative:

\[ \frac{d^2S}{dt^2} = \frac{d}{dt}(t^2 - 4t + 3) = 2t - 4 \]

### Step 5: Evaluate the second derivative at the critical points
We will evaluate the second derivative at the critical points \( t = 1 \) and \( t = 3 \):

1. For \( t = 1 \):

\[ \frac{d^2S}{dt^2} \bigg|_{t=1} = 2(1) - 4 = 2 - 4 = -2 \quad (\text{Negative, indicating a maximum}) \]

2. For \( t = 3 \):

\[ \frac{d^2S}{dt^2} \bigg|_{t=3} = 2(3) - 4 = 6 - 4 = 2 \quad (\text{Positive, indicating a minimum}) \]

### Conclusion
- The function \( S(t) \) has a **maximum** at \( t = 1 \).
- The function \( S(t) \) has a **minimum** at \( t = 3 \).