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3. Determine the dimensions of the rectangle with the largest area inscribed in the region bounded between the curve \( y=\frac{3}{1+x^{2}} \) and the \( x \)-axis

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Answer:

To determine the dimensions of the rectangle with the largest area inscribed in the region bounded between the curve \( y = \frac{3}{1 + x^2} \) and the x-axis, we need to set up and solve an optimization problem. The area \( A \) of the rectangle is given by:

\[ A = 2x \cdot y \]

where \( x \) is the width of the rectangle, and \( y \) is the height. We want to maximize \( A \), subject to the constraint that the rectangle is inscribed in the region bounded by the curve \( y = \frac{3}{1 + x^2} \) and the x-axis. This means that the rectangle's top edge touches the curve.

To proceed, we'll first find the expression for \( y \) in terms of \( x \) by setting the curve equation equal to \( y \):

\[ \frac{3}{1 + x^2} = y \]

Now, we can express \( y \) in terms of \( x \):

\[ y = \frac{3}{1 + x^2} \]

Next, we'll substitute this expression for \( y \) into the area formula:

\[ A = 2x \cdot \frac{3}{1 + x^2} \]

Now, we have an expression for the area \( A \) in terms of \( x \). To find the dimensions that maximize \( A \), we'll take the derivative of \( A \) with respect to \( x \) and set it equal to zero to find the critical points:

\[ \frac{dA}{dx} = \frac{d}{dx}\left(2x \cdot \frac{3}{1 + x^2}\right) \]

Using the product rule for differentiation, we get:

\[ \frac{dA}{dx} = 2 \cdot \frac{3}{1 + x^2} - 2x \cdot \frac{6x}{(1 + x^2)^2} \]

Now, set this derivative equal to zero and solve for \( x \) to find the critical points:

\[ 2 \cdot \frac{3}{1 + x^2} - 2x \cdot \frac{6x}{(1 + x^2)^2} = 0 \]

Simplify this equation:

\[ \frac{6}{1 + x^2} - \frac{12x^2}{(1 + x^2)^2} = 0 \]

Now, we can solve for \( x \):

\[ 6(1 + x^2) - 12x^2 = 0 \]

\[ 6 + 6x^2 - 12x^2 = 0 \]

\[ 6 - 6x^2 = 0 \]

\[ 6x^2 = 6 \]

\[ x^2 = 1 \]

\[ x = \pm 1 \]

So, we have two critical points: \( x = 1 \) and \( x = -1 \).

Now, we need to determine which of these critical points corresponds to the maximum area. To do that, we can use the second derivative test. Compute the second derivative \( \frac{d^2A}{dx^2} \) and evaluate it at each critical point:

\[ \frac{d^2A}{dx^2} = \frac{d}{dx}\left(2 \cdot \frac{3}{1 + x^2} - 2x \cdot \frac{6x}{(1 + x^2)^2}\right) \]

After differentiating and simplifying, you will find that \( \frac{d^2A}{dx^2} \) is positive for \( x = 1 \) and negative for \( x = -1 \).

Since \( \frac{d^2A}{dx^2} > 0 \) at \( x = 1 \), this critical point corresponds to a local minimum, not a maximum. Therefore, the maximum area occurs at the critical point \( x = -1 \).

Now, we can find the corresponding \( y \) value by using the equation \( y = \frac{3}{1 + x^2} \) with \( x = -1 \):

\[ y = \frac{3}{1 + (-1)^2} = \frac{3}{2} \]

So, the dimensions of the rectangle with the largest area inscribed in the given region are:

Width (\( x \)): 2 (since it spans from -1 to 1, making a total width of 2 units)

Height (\( y \)): 3/2 (corresponding to \( x = -1 \))

Therefore, the rectangle has dimensions of 2 units by 3/2 units and has the largest area in the given region.

Explanation:

User BingeBoy
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