Let a parabola be `y=12-x^2`. Find the maximum area of rectangle whose base lie on x-axis and two points lie on parabola. (A) `8` (B) `4` (C) `32` (D) `34`

Options

A. `20sqrt2`
B. `18sqrt2`
C. 32
D. 36

Answer

To find the maximum area of a rectangle whose base lies on the x-axis and whose upper corners lie on the parabola \(y = 12 - x^2\), we can follow these steps:

### Step 1: Define the rectangle
Let the points where the rectangle touches the parabola be \((\alpha, y)\) and \((- \alpha, y)\). The height of the rectangle will be \(y\) and the width will be \(2\alpha\).

### Step 2: Express the height in terms of \(\alpha\)
From the equation of the parabola, we can express \(y\) in terms of \(\alpha\):
\[ y = 12 - \alpha^2 \]

### Step 3: Write the area of the rectangle
The area \(A\) of the rectangle can be expressed as:
\[ A = \text{base} \times \text{height} = (2\alpha)(12 - \alpha^2) \]
Thus,
\[ A = 2\alpha(12 - \alpha^2) = 24\alpha - 2\alpha^3 \]

### Step 4: Differentiate the area with respect to \(\alpha\)
To find the maximum area, we need to differentiate \(A\) with respect to \(\alpha\):
\[ \frac{dA}{d\alpha} = 24 - 6\alpha^2 \]

### Step 5: Set the derivative to zero to find critical points
Setting the derivative equal to zero to find critical points:
\[ 24 - 6\alpha^2 = 0 \]
\[ 6\alpha^2 = 24 \]
\[ \alpha^2 = 4 \]
\[ \alpha = 2 \quad (\text{since } \alpha \text{ must be positive}) \]

### Step 6: Calculate the maximum area
Now, substitute \(\alpha = 2\) back into the area formula:
\[ A = 2(2)(12 - 2^2) = 4(12 - 4) = 4 \times 8 = 32 \]

### Conclusion
The maximum area of the rectangle is \(32\).

### Final Answer
Thus, the answer is \( \boxed{32} \).

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