Problem 53 - Essential Calculus

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.

Step-by-Step Solution

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 1: 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 \(A(x) = 2x*e^{-x^{2}}\).

Step 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 \(A'(x) = 2e^{-x^{2}} - 4x^{2}e^{-x^{2}}\).

Step 3: 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 \(1 - 2x^{2} = 0\).

Step 4: 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 \(x = \pm 1/\sqrt{2}\).

Step 5: 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 \(x = 1/\sqrt{2}\).

Step 6: 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.

Key Concepts

Area under a curve

To find the largest rectangle that can be inscribed under a curve, it's helpful to understand the concept of the area under a curve. Imagine the function described by the curve, which in this case is given by the equation \(y=e^{-x^{2}}\). This exponential function forms a bell-shaped curve across the graph. The goal is to find the rectangle with the largest area that sits perfectly under this curve within the first and second quadrants. To achieve this, we imagine drawing rectangles with varying widths and heights beneath the curve. The height of the rectangle is determined by the function value at a given \(x\)-coordinate, while the width extends from \(-x\) to \(x\) since it exists in both the left and right quadrants. We calculate this area to find the optimal dimensions of such a rectangle, effectively making us explore how widths and heights need to change to find the largest possible area that fits below the curve.

Product Rule

The product rule is an essential calculus concept used to find derivatives when two functions are multiplied together, as seen in the expression \(A(x) = 2x \, e^{-x^{2}}\). This expression represents the area function of our rectangle. Notice that it's the product of \(2x\) and \(e^{-x^{2}}\), hence to find its derivative, we apply the product rule. The product rule states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product \(u(x)v(x)\) is:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]Using this formula allows us to differentiate \(A(x)\). It helps identify how the area changes as the width \(x\) of the rectangle changes, essential for optimizing the rectangle’s dimensions.

Derivative Test

The derivative test is a crucial method for finding maxima and minima in calculus, particularly when optimizing functions. After differentiating the area function \(A(x) = 2x \, e^{-x^{2}}\) with respect to \(x\), we employ the first derivative test to locate the maximum area. Here's how it works:- Set the derivative \(A'(x)\) equal to zero to find critical points.- This reveals \(A'(x) = 2e^{-x^{2}} - 4x^{2}e^{-x^{2}} = 0\).Factoring out common terms and solving gives feasible \(x\)-values of \(\pm\frac{1}{\sqrt{2}}\) for the potential maxima. - Apply the first derivative test by evaluating the sign changes around these critical points.- The sign change of \(A'(x)\) informs us that the maximum area occurs at these values, meaning it transitions from positive to negative or vice versa.This method ensures we correctly identify the \(x\)-values that maximize the rectangle area under the curve.

Exponential Function

Exponential functions are a fundamental concept in calculus, often appearing in problems involving growth and decay. In our exercise, the exponential function \(e^{-x^{2}}\) dictates the curve beneath which we seek to inscribe our rectangle. Here's why it matters:- The exponential function \(e^{-x^{2}}\) doesn't equal zero anywhere, which influences our optimization process.- This is important when solving for \(x\) as part of finding the maximum area.Exponential decreases quickly as \(x\) moves away from zero, creating the bell curve shape. This steep decline implies that larger rectangle areas come from central part close to \(x = 0\). When analyzing problems like these, understanding how exponential functions behave helps us deduce the characteristics of the optimal solutions in areas under their curves.