Problem 35 - Calculus of a Single Variable

Question

Minimum Area The sum of the perimeters of an equilateral triangle and a square is \(10 .\) Find the dimensions of the triangle and the square that produce a minimum total area.

Step-by-Step Solution

Answer

The dimensions that yield a minimum total area are \(b = \frac{30(4 - \sqrt{3})}{25}\) for the triangle and \(a = \frac{10 \sqrt{3} - 6}{10}\) for the square.

Step 1: Set up the Equations

Let's denote the side of the square by \(a\) and that of the equilateral triangle by \(b\). The fact that the sum of the perimeters equals ten can be written as:\[4a + 3b = 10\]Furthermore, we can express the total area of both figures as:\[A = a^2 + \frac{\sqrt{3}}{4}b^2\]

Step 2: Derive an Expression for \(a\)

Let's solve the perimeter equation for \(a\), which yields:\[a = \frac{10 - 3b}{4}\]

Step 3: Combine and Simplify

Let's substitute \(a\) into \(A\), which leads to:\[A = \left(\frac{10 - 3b}{4}\right)^2 + \frac{\sqrt{3}}{4}b^2\]Simplify the equation to form:\[A = \frac{25b^2 - 60b + 100}{16} + \frac{\sqrt{3}}{4}b^2\]

Step 4: Differentiate w.r.t \(b\)

We can minimise this by finding where its derivative is zero:\[A' = \frac{50b - 60}{16} + \frac{\sqrt{3}}{2}b\]Setting the derivative equal to zero, we have:\[\frac{50b - 60}{16} + \frac{\sqrt{3}}{2}b = 0\]

Step 5: Solve for \(b\)

Solving the equation will give:\[b = \frac{30(4 - \sqrt{3})}{25}\]

Step 6: Solve for \(a\)

Substitute this into the equation for \(a\) to receive:\[a = \frac{10 - 3b}{4} = \frac{10 \sqrt{3} - 6}{10}\]

Step 7: Verify the Result

We could also check the result by calculating the second derivative of A and confirming that it's positive for our result \(b\), thus indicating a local minimum. For our case, the computed \(a\) and \(b\) result in a minimum total area.

Key Concepts

Perimeter of Geometric Shapes

Understanding the perimeter of geometric shapes is a crucial concept in optimization problems. The perimeter is the total distance around the boundary of a shape. It's essentially the length of the outline of a shape. In the context of our exercise, the focus is on two shapes: an equilateral triangle and a square.
For a square, the perimeter is calculated by adding up the lengths of its four equal sides, which can be expressed as the equation \(4a\), where \(a\) is the side length of the square. Similarly, the perimeter of an equilateral triangle, which has three equal sides, is given by \(3b\), where \(b\) is the side length of the triangle. In our optimization problem, the sum of the perimeters of these two figures is given as 10, which can be represented by the equation \(4a + 3b = 10\). This constraint relates the side lengths of the two shapes and is an essential stepping stone in finding the dimensions that minimize the total area.
By obtaining a relationship between the two variables, in this case \(a\) and \(b\), we can use it to express one variable in terms of the other, simplifying the problem and allowing for further analysis through differentiation, which will be discussed in a later section.

Area of Geometric Shapes

In addition to the perimeter, understanding how to calculate the area of geometric shapes is fundamental in solving optimization problems. The area measures the extent of a two-dimensional shape or surface and is expressed in square units.
In the given exercise, we consider the area of a square and an equilateral triangle. The area of a square is found by squaring its side length \(a\), denoted as \(a^2\). For an equilateral triangle, the formula is more intricate due to its geometry, and the area can be expressed as \(\frac{\sqrt{3}}{4}b^2\), where \(b\) is the side length.
In our problem, we aim to find the minimum combined area of both the triangle and the square. The total area can therefore be represented as \(A = a^2 + \frac{\sqrt{3}}{4}b^2\). It's important to understand these area formulas to set up the next steps for optimization, which involve combining these areas into a single equation that only depends on one variable, and eventually finding this variable's optimum value.

Differentiation for Minima

Differentiation is an essential tool in calculus used to find the rate at which a function is changing at any given point. This becomes exceedingly useful in optimization problems, especially when seeking to determine minima or maxima of functions.
In the context of our problem, once we've expressed the total area \(A\) in terms of a single variable by substituting the relationship between \(a\) and \(b\) from the perimeter constraint, we can then differentiate this area function with respect to that variable. Here, we differentiate our expressed area function with respect to \(b\), yielding a new function \(A'\).
By finding where this derivative equals zero, we can identify critical points, which could potentially be the minima we are searching for. However, a zero derivative alone does not guarantee a minimum; we must also ensure that the second derivative is positive at this critical point, confirming it's indeed a local minimum and not a maximum or inflection point.
To summarize, the application of differentiation in our problem involves finding the first derivative with respect to \(b\), setting it equal to zero to find critical points, and then using the second derivative test to confirm the nature of these points. This step-by-step process leads us to the dimensions of the triangle and square that produce a minimum total area.