Find the points of absolute maximum and minimum of each of the following :
`y=(1)/(3)x^(3//2)-4x,0lexle64`

Answer

To find the points of absolute maximum and minimum for the function \( y = \frac{1}{3}x^{\frac{3}{2}} - 4x \) over the interval \( [0, 64] \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function with respect to \( x \):

\[ y' = \frac{d}{dx}\left(\frac{1}{3}x^{\frac{3}{2}} - 4x\right) \]

Using the power rule, we get:

\[ y' = \frac{1}{3} \cdot \frac{3}{2} x^{\frac{3}{2} - 1} - 4 = \frac{1}{2} x^{\frac{1}{2}} - 4 \]

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

\[ \frac{1}{2} x^{\frac{1}{2}} - 4 = 0 \]

Solving for \( x \):

\[ \frac{1}{2} x^{\frac{1}{2}} = 4 \]
\[ x^{\frac{1}{2}} = 8 \]
\[ x = 64 \]

### Step 3: Evaluate the function at the endpoints and critical points
Now, we will evaluate the function at the endpoints \( x = 0 \) and \( x = 64 \), as well as at the critical point \( x = 64 \).

1. **At \( x = 0 \)**:
\[ y(0) = \frac{1}{3}(0)^{\frac{3}{2}} - 4(0) = 0 \]

2. **At \( x = 64 \)**:
\[ y(64) = \frac{1}{3}(64)^{\frac{3}{2}} - 4(64) \]
First, calculate \( (64)^{\frac{3}{2}} = (8^2)^{\frac{3}{2}} = 8^3 = 512 \):
\[ y(64) = \frac{1}{3}(512) - 256 = \frac{512}{3} - 256 \]
Converting \( 256 \) to a fraction:
\[ 256 = \frac{768}{3} \]
Thus:
\[ y(64) = \frac{512 - 768}{3} = \frac{-256}{3} \]

### Step 4: Compare the values to find absolute maximum and minimum
Now we compare the values obtained:

- \( y(0) = 0 \)
- \( y(64) = -\frac{256}{3} \)

The absolute maximum value occurs at \( x = 0 \) with \( y = 0 \), and the absolute minimum value occurs at \( x = 64 \) with \( y = -\frac{256}{3} \).

### Final Answer
- Absolute maximum point: \( (0, 0) \)
- Absolute minimum point: \( (64, -\frac{256}{3}) \)

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