Question

Find the area of the largest rectangle that can be inscribed under the curve \(y=e^{-x^{2}}\) in the first and second quadrants.

Short answer

The maximum possible area of the rectangle that can be inscribed under the curve \(y=e^{-x^{2}}\) in the first and second quadrants is \( \sqrt{2}*e^{-(1/2)}\).

Step-by-step solution

Write the area function in terms of one variable

Since \(y=e^{-x^{2}}\) and \(A = 2xy = 2x*e^{-x^{2}}\), the area \(A\) as a function of \(x\) is \(A(x) = 2x*e^{-x^{2}}\).

Find the derivative of the area function

We take the derivative of \(A(x)\) with respect to \(x\) using the product rule, which results in \(A'(x) = 2e^{-x^{2}} - 4x^{2}e^{-x^{2}}\).

Set the derivative equal to zero

We then set the derivative equal to zero and solve for \(x\) : \(2e^{-x^{2}} - 4x^{2}e^{-x^{2}} = 0\). Factoring out \(2e^{-x^{2}}\) gives \(2e^{-x^{2}}(1 - 2x^{2}) = 0\). This equation is solved if \(e^{-x^{2}} = 0\) or \(1 - 2x^{2} = 0\).

Solve for x

The exponential function is never zero, so the first part \(e^{-x^{2}} = 0\) has no solution. The second part \(1 - 2x^{2} = 0\) gives \(x^{2} = 1/2\), so \(x = \pm 1/\sqrt{2}\).

First derivative test

We use the first derivative test to determine which of these values gives the maximum area. \(A'(x) = 2e^{-x^{2}} - 4x^{2}e^{-x^{2}}\) changes sign from positive to negative at \(x = -1/\sqrt{2}\) and from negative to positive at \(x = 1/\sqrt{2}\), so the area \(A\) has a maximum at \(x = -1/\sqrt{2}\) and \(x = 1/\sqrt{2}\).

Find the maximum area

The maximum area is obtained by substituting \(x = 1/\sqrt{2}\) or \(x = -1/\sqrt{2}\) (since we are looking in the first and second quadrant, both will give the same area) into the area function, it results in \(A = 2*(1/\sqrt{2})*e^{-(1/2)} = sqrt{2}*e^{-(1/2)} – this is the maximum area of the rectangle.