Problem 13 - Calculus Single Variable

Question

Solve each of the maximum-minimum problems. Some may not have a solution, whereas others may have their solution at the endpoint of the interval of definition. A rectangle has its base on the \(x\) -axis, and its upper corners are on the graph of \(y=4-x^{2} .\) What dimensions for the rectangle will give it maximal area?

Step-by-Step Solution

Answer

Width = \(2\sqrt{\frac{4}{3}}\), Height = \(\frac{8}{3}\), for maximum area.

Step 1: Define the problem

We need to find the dimensions of a rectangle with its base on the \(x\)-axis and its upper corners on the parabola \(y = 4-x^2\), so as to maximize the area. The width of the rectangle will be \(2x\) and the height will be \(y\). So the area \(A\) of the rectangle can be expressed as a function of \(x\): \(A(x) = 2x \cdot y = 2x(4 - x^2)\).

Step 2: Express the area as a function of x

Substitute \(y = 4-x^2\) into the expression for area: \[ A(x) = 2x(4 - x^2) = 8x - 2x^3. \]Now, we have a function \(A(x)\) that we need to maximize.

Step 3: Find the derivative of the area function

Determine \(A'(x)\) to find critical points where the area function is maximized or minimized:\[ A'(x) = \frac{d}{dx}(8x - 2x^3) = 8 - 6x^2. \]

Step 4: Find the critical points

Set the derivative \(A'(x) = 0\) to find critical points:\[ 8 - 6x^2 = 0 \]Solving for \(x\), we get:\[ 6x^2 = 8 \]\[ x^2 = \frac{4}{3} \]\[ x = \pm \sqrt{\frac{4}{3}}. \]

Step 5: Evaluate endpoints and critical points

Check the value of the area at endpoints and critical points. The domain for \(x\) is between \(-2\) and \(2\) because the parabola intersects the \(x\)-axis at these points.1. At \(x = \sqrt{\frac{4}{3}}\) (only consider positive as the rectangle width cannot be negative):\[ A\left(\sqrt{\frac{4}{3}}\right) = 8\left(\sqrt{\frac{4}{3}}\right) - 2\left(\sqrt{\frac{4}{3}}\right)^3 \]\[ A\left(\sqrt{\frac{4}{3}}\right) = 8\sqrt{\frac{4}{3}} - \frac{8\sqrt{\frac{4}{3}}}{3} = \frac{16\sqrt{3}}{3}. \]2. At \(x = -2\) and \(x = 2\) (the area is zero at endpoints, as the rectangle collapses):\(A(-2) = A(2) = 0.\)

Step 6: Conclusion about maximal area

The maximum area occurs at \(x = \sqrt{\frac{4}{3}}\) and at this point, the dimensions of the rectangle are: \(width = 2\sqrt{\frac{4}{3}}\) and \(height = 4 - \left(\sqrt{\frac{4}{3}}\right)^2 = \frac{8}{3}. \) Therefore, the dimensions \(2\times \sqrt{\frac{4}{3}}\times \frac{8}{3}\) give the maximum area.

Key Concepts

Critical Points

In optimization problems, such as finding the maximum area of a geometric figure, we often have to identify points called critical points. These are the points in the domain of a function where the derivative is zero or undefined. These points are particularly important because they may represent local maxima, minima, or points of inflection. In our problem, critical points help us figure out where the rectangle might have a maximum area.

To find the critical points, we begin by taking the derivative of the function that represents the area. Setting the derivative to zero enables us to locate where any potential extrema might be. While only solving for when the derivative is zero might not give a complete picture, it is a pivotal step in identifying where to investigate further for maximum or minimum values.

Derivative

The derivative plays a crucial role in optimization problems by helping us determine the rate of change of a function. Essentially, it tells us how the output or area of the rectangle changes as we adjust the dimensions. In our example, the area function is given by \(A(x) = 8x - 2x^3\).

Calculating the derivative \(A'(x) = 8 - 6x^2\) helps identify where the rate of change is zero, leading us to possible maximum or minimum values. The derivative enables us to not only find critical points but also to understand the nature of these points by examining the sign of the derivative before and after these points. This tells us whether the function is increasing or decreasing, thus aiding our analysis of the problem.

Maximum Area

In optimization, one of the main goals is to find the maximum area or volume, depending on the problem context. For our rectangle problem, the maximum area is what we're after. Knowing both the function \(A(x) = 8x - 2x^3\) and its derivative \(A'(x) = 8 - 6x^2\), enables us to precisely identify which parts of the domain produce the largest output or area.

After finding the critical points by setting \(A'(x) = 0\), we assess the area at these points along with the endpoints of the domain. The endpoints might reveal maximum or minimum values as well. In this case, analysis showed that the peak area occurs at \(x = \sqrt{\frac{4}{3}}\), where the function achieves its largest value distinct from zero.

Parabola

A parabola is a symmetrical, u-shaped curve that is the graph of a quadratic function like \(y = 4 - x^2\), as seen in our example. In optimization problems, light is often shed on the problem by understanding the shape and behavior of any intersecting curves with the object of interest, here being the rectangle.

The parabola plays the role of the upper boundary for the rectangle. The function \(y = 4 - x^2\) tells us where the edges of the rectangle can stretch to vertically. The curve helps in determining the feasible domain for our variable \(x\), which extends between points where the parabola intersects the \(x\)-axis: at \(-2\) and \(2\). This sets our boundaries for search for the maximum area, ensuring the rectangle never overextends beyond physical or mathematical reason.