Question

Four feet of wire is to be used to form a square and a circle. (a) Express the sum of the areas of the square and the circle as a function \(A\) of the side of the square \(x\). (b) What is the domain of \(A\) ? (c) Use a graphing utility to graph \(A\) on its domain. (d) How much wire should be used for the square and how much for the circle in order to enclose the least total area? the greatest total area?

Short answer

The total area of the square and the circle can be expressed as \(A(x)=x^{2}+((4-4x)^2)/(4\pi)\) where \(0<x<1\). For the least total area, more wire should be allocated to the square than the circle, and for the greatest total area, all the wire should be allocated to the circle.

Step-by-step solution

Understand the problem and use the given data

The problem is to form a square and a circle from 4 feet of wire. The perimeter of a square is \(4x\), where \(x\) represents the length of the side of a square, and the circumference of a circle is \(2\pi r\), where \(r\) represents the radius of the circle. We have the total wire length as 4. So, we can write an equation as: \(4x+2\pi r = 4\).

Express \(\pi r\) in terms of \(x\)

From the equation \(4x+2\pi r = 4\), the circumference of the circle (i.e., \(2\pi r\)) can be expressed as \(4-4x\). Divide by \(2\pi\) to get \(r=(4-4x)/(2\pi)\).

Write the function \(A(x)\) for the areas of the square and the circle

The area of the square can be represented as\(x^2\) and the area of the circle is \(\pi r^2\), which can be writtenas \(\pi ((4-4x)/(2\pi))^2\). This gives the area function as \(A(x)= x^2+ ((4-4x)^2)/(4\pi)\) with the domain \(01\). This is because \(x\) can't be zero as it potential for the circle consuming all the wire, and can't be higher than 1 as potently for the square consuming all the wire.

Find the wire lengths for the least and greatest enclosed areas

You could use calculus or a graphing utility to determine the minimum and maximum of \(A(x)\) within its domain. For the least enclosed area (minimum of \(A(x)\)), the wire should be allocated such that the square receives more length than the circle, as squares occupy more area with less perimeter than circles. For the greatest enclosed area (maximum of \(A(x)\)), the wire should be allocated such that the circle receives all the length, as circles occupy less area with more circumference than squares.

Key concepts

Area of a Square and a Circle

Understanding how to calculate the area of squares and circles is fundamental in geometry. For a square, the area is found by squaring the length of one of its sides. This is because a square has all sides of equal length, and its area is simply a side length \times itself, represented by the formula, \( A_{\text{square}} = x^2 \), where \( x \) is the side length.

As for a circle, the area is linked to its radius. The further from the center, the more space it covers, which is why we use the formula \( A_{\text{circle}} = \(\pi r^2 \)\), with \( r \) being the radius. Here, \( \pi \) represents the mathematical constant approximately equal to 3.14159, which expresses the ratio of the circumference of any circle to its diameter. When optimizing wire lengths to make both a square and a circle, careful calculations using these formulas helps in determining how much wire to allocate to each shape to either maximize or minimize the total area.

Function of a Variable

When dealing with functions in mathematics, it’s imperative to understand the term 'variable'. In the context of our exercise, we define a function \( A(x) \) to describe the total area created by both the square and the circle. The variable \( x \) represents the side length of the square, which is pivotal in this scenario because the amount of wire used for the square directly affects the radius—and consequently, the area—of the circle.

Thus, the function \( A(x) \) encapsulates the relationship between the side length of the square and the total area covered by both shapes. It offers a systematic way to determine how changing \( x \) influences the outcome, which in this case, is the combined area. Defining such a function is a powerful tool, as it allows you to investigate different scenarios by simply substituting different values for \( x \) to see how they affect the total area.

Domain of a Function

The domain of a function is essentially the set of all possible input values (values for the independent variable) for which the function is defined and will produce a valid output. In our case, the domain of \( A(x) \) is the set of all side lengths \( x \) that are feasible given the wire length constraints. As mentioned, \(x\) can't be zero, as it would imply no square, nor can it be greater than 1, as this would require more wire than available.

Thus, the domain for \( x \) is more precisely expressed as: \( 0 < x < 1 \). It represents all potential side lengths of the square that can be formed without exceeding the total wire length of 4 feet. When solving such problems, determining the domain is critical as it restricts the values that \( x \) can take and ensures the solution is practical and relevant to the given context.

Extrema of a Function

In mathematics, the extrema (plural of extremum) of a function are the points in its domain where the function takes minimum or maximum values. These concepts are pivotal in optimization problems such as the one considered, where we want to find the allocation of wire that results in the least or greatest total area for our shapes.

To locate these points, one could use calculus techniques, specifically finding the derivative of the function \( A(x) \) and setting it to zero to figure out the critical points, after which you would determine whether these points represent minima or maxima. Another way is to graph the function, as visually identifying these extrema can be more intuitive. For the minimum area, the wire should be divided such that more of it is used for the square, since a square encloses more area per unit of perimeter compared to a circle. On the contrary, for maximum area, allocating all the wire for the circle would be optimal, given the wire length of 4 feet.