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Chapter 5: Problem 73

Area 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

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

Step by step solution

01

Define the Area

Let one side of the rectangle be along the x-axis and the other side be along the y-axis. Then the top of the rectangle will touch the curve at some positive \(x_0\) and negative \(x_0\). Given that \(y = e^{-x^2}\), the height of the rectangle is \(e^{-x_0^2}\) and the width is \(2x_0\). Therefore, the area of the rectangle, A, can be written as a function of \(x_0\): \(A(x_0) = 2x_0e^{-x_0^2}\).
02

Optimize the Function

To find the maximum of the function, take the derivative \(A'(x_0)\) and equate it to 0, and solve for \(x_0\). \(A'(x_0)= 2e^{-x_0^2}(1-2x_0^2)\). Solving this equation gives \(x_0 = 0\) and \(x_0 = \pm \frac{1}{\sqrt{2}}\).
03

Check the Second Derivative

To confirm the maximum, we can take the second derivative and analyze its signs. The second derivative is \(A''(x_0)= -4e^{-x_0^2}(2x_0^2-1)\). Plugging in \(x_0 = 0\), we get \(A''(0) = 4 > 0\), which is indicative of a minimum. For other \(x_0 = \pm \frac{1}{\sqrt{2}}\), we get \(A''(-\frac{1}{\sqrt{2}}) = A''(\frac{1}{\sqrt{2}}) = -2e^{-1} < 0\). Hence, we get maximum area at \(x_0 = \pm \frac{1}{\sqrt{2}}\).
04

Find the Maximum Area

The maximum area A is found by plugging in \(x_0 = \frac{1}{\sqrt{2}}\) or \(x_0 = -\frac{1}{\sqrt{2}}\) into the area function \(A(x_0)\): \(A(-\frac{1}{\sqrt{2}}) = A(\frac{1}{\sqrt{2}}) = 2(\frac{1}{\sqrt{2}})e^{-\frac{1}{2}} = \sqrt{2e^{-1}}\).

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