Find the area of the largest rectangle in the first quadrant with two sides on x-axis and y-axis and one vertex on the curve `y= 12-x^2`

Answer

To find the area of the largest rectangle in the first quadrant with two sides on the x-axis and y-axis and one vertex on the curve \( y = 12 - x^2 \), we can follow these steps:

### Step 1: Understand the problem
We need to find a rectangle in the first quadrant where one vertex lies on the curve \( y = 12 - x^2 \). The rectangle will have its base along the x-axis and its height along the y-axis.

### Step 2: Define the rectangle's dimensions
Let the vertex of the rectangle on the curve be at the point \( (x, y) \). The rectangle will then have:
- Length = \( x \) (along the x-axis)
- Height = \( y \) (along the y-axis)

### Step 3: Express the area of the rectangle
The area \( A \) of the rectangle can be expressed as:
\[ A = \text{Length} \times \text{Height} = x \times y \]
Since the vertex lies on the curve, we can substitute \( y \) with \( 12 - x^2 \):
\[ A = x(12 - x^2) = 12x - x^3 \]

### Step 4: Differentiate the area function
To find the maximum area, we need to differentiate the area function \( A \) with respect to \( x \):
\[ \frac{dA}{dx} = 12 - 3x^2 \]

### Step 5: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 12 - 3x^2 = 0 \]
Solving for \( x \):
\[ 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2 \]
Since we are in the first quadrant, we take \( x = 2 \).

### Step 6: Find the corresponding \( y \)
Now, substitute \( x = 2 \) back into the equation for \( y \):
\[ y = 12 - x^2 = 12 - 2^2 = 12 - 4 = 8 \]

### Step 7: Calculate the area
Now we can calculate the area of the rectangle:
\[ A = x \times y = 2 \times 8 = 16 \]

### Conclusion
The area of the largest rectangle is \( 16 \) square units.