Problem 10 - Calculus Early Transcendentals

Question

Find the point \(P\) in the first quadrant on the curve \(y=x^{-2}\) such that a rectangle with sides on the coordinate axes and a vertex at \(P\) has the smallest possible perimeter.

Step-by-Step Solution

Answer

Point \(P\) is \( \left( 2^{1/3}, 2^{-2/3} \right) \).

Step 1: Express Perimeter Function

Let the point on the curve be \((x, y)\) where \(y = x^{-2}\). The rectangle with a vertex at this point will have sides on the x-axis and y-axis. Therefore, the perimeter \(P\) of the rectangle is given by \(2x + 2y\). Substituting for \(y\), we have the perimeter function: \[ P = 2x + 2x^{-2} \]

Step 2: Differentiate the Perimeter Function

To find the minimum perimeter, we need to differentiate the perimeter function with respect to \(x\). \[ \frac{dP}{dx} = 2 - 4x^{-3} \]

Step 3: Find Critical Points

Set the derivative equal to zero to find critical points: \[ 2 - 4x^{-3} = 0 \]Solving for \(x\), \[ 4x^{-3} = 2 \]\[ x^{-3} = \frac{1}{2} \]\[ x = \left(\frac{1}{2}\right)^{-1/3} = 2^{1/3} \]

Step 4: Verify Minimum

To ensure that this critical point is a minimum, we check the second derivative: \[ \frac{d^2P}{dx^2} = 12x^{-4} \]For \(x > 0\), \(12x^{-4} > 0\), which confirms it is a minimum.

Step 5: Find the Coordinates of Point P

Substitute \(x = 2^{1/3}\) back into the expression for \(y\) to find \(y\). \[ y = (2^{1/3})^{-2} = 2^{-2/3} \]Thus, the point \(P\) is \( \left( 2^{1/3}, 2^{-2/3} \right) \).

Key Concepts

Perimeter Function

To solve optimization problems in calculus, setting up an appropriate function is crucial. In this exercise, we're dealing with a geometric shape — a rectangle — derived from a curve, and we aim to minimize its perimeter. A perimeter function essentially measures the total length around a shape. For rectangles, the perimeter (P) is given as twice the sum of its length and width. In this case, imagine a rectangle's vertex at point \((x, y)\) on the curve \(y = x^{-2}\), where the adjacent sides align with the coordinate axes. Thus, the rectangle's dimensions are determined directly by the coordinates: one side is \(x\) and the other is \(y\).
Substituting \(y = x^{-2}\) into the perimeter equation, we get the function \(P = 2x + 2y\). The significance of this substitution simplifies our task to purely optimizing against \(x\), giving us the formula:

• \(P(x) = 2x + 2x^{-2}\)

Understanding the perimeter function is the first step towards finding where it achieves its smallest value.

Differentiation

Differentiation is an essential tool in optimization problems. It involves finding the derivative of a function, which measures the rate of change of that function with respect to a variable, typically \(x\). In this context, we want to understand how the perimeter of our rectangle changes as \(x\) changes. We do this by differentiating our perimeter function \(P(x) = 2x + 2x^{-2}\) with respect to \(x\).
By applying the rules of differentiation:

• The derivative of \(2x\) is simply \(2\).
• For \(2x^{-2}\), applying the power rule, we get \(-4x^{-3}\).

So, the derivative is:

• \(\frac{dP}{dx} = 2 - 4x^{-3}\)

The resulting expression tells us how the perimeter is changing at any given \(x\), which is critical for finding minimum points.

Critical Points

Once the derivative of a function is determined, the next step focuses on finding critical points. Critical points are where the derivative is zero or undefined, these are potential points of minimum or maximum values of the function. For the perimeter function derivative \(\frac{dP}{dx} = 2 - 4x^{-3}\), we set this equal to zero to locate critical points:

• \(2 - 4x^{-3} = 0\)
• \(4x^{-3} = 2\)
• \(x^{-3} = \frac{1}{2}\)
• Solving gives us \(x = \left(\frac{1}{2}\right)^{-1/3} = 2^{1/3}\)

This solution provides the x-coordinate of the point where the perimeter might be minimized. By identifying this point, we're a step closer to optimizing the rectangle's perimeter.

Second Derivative Test

With the critical point identified, confirming whether it's a minimum is crucial. The second derivative test is used in calculus to determine the concavity of a function at a given point, which tells us if the point is a minimum or maximum. Essentially:

• If the second derivative at a point is positive, the function is concave up, indicating a local minimum.
• If negative, it's concave down, indicating a local maximum.

For our perimeter function \(P(x) = 2x + 2x^{-2}\), the second derivative is:

• \(\frac{d^2P}{dx^2} = 12x^{-4}\).

At \(x = 2^{1/3}\), substituting in gives a positive value:

• \(12x^{-4} > 0\)

Hence, the point \((x, y) = \left(2^{1/3}, \left(2^{1/3}\right)^{-2}\right)\) is indeed a minimum point, ensuring the rectangle has the smallest possible perimeter at this configuration.