Question

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

Short answer

The dimensions of the square and triangle that will minimize the total area are approximately \(1.7207\) for the square and \(0.8539\) for the triangle.

Step-by-step solution

Express the Given Condition

Denote the side of the square as \(x\) and one side of the equilateral triangle as \(y\). According to the condition, we have \(4x+3y=10\). This can be rearranged to express (\(y\)) in terms of (\(x\)): \(y=(10-4x)/3\).

Express the Target Function

The area of the square is \(A_s=x^2\) and the area of the equilateral triangle is \(A_t = y^2\sqrt{3}/4 = ((10-4x)^2\sqrt{3}/36\). The total area is \(A = A_s + A_t = x^2 + ((10-4x)^2\sqrt{3}/36\).

Derive the Function and Find its Zeroes

To find the minimum, we derive the total area with respect to \(x\): \(A' = 2x - ((10-4x) \cdot 8 \sqrt{3}/108)=0\). Solving this equation for \(x\) yields: \(x \approx 1.7207\) and \(y= (10-4x)/3 \approx 0.8539\).

Verify the Minimum

To confirm that this solution indeed yields a minimum, we compute the second derivative of \(A\), \(A''=2+32\sqrt{3}/108>0\). Since the second derivative is positive, the area is indeed minimized.

Key concepts

Perimeter of Geometrical Figures

Understanding the perimeter is fundamental in optimizing shapes for various conditions. The perimeter is the total length around a two-dimensional shape. For instance, to calculate the perimeter of a square, you would add up the lengths of all four sides. Since all four sides are equal, this can be simplified to multiplying the length of one side by four, resulting in the formula:
\( P_{square} = 4 \times side \).
In contrast, an equilateral triangle, which has three sides of equal length, would have a perimeter calculated by tripling one side's length, as follows:
\( P_{triangle} = 3 \times side \).
When given the sum of perimeters from different shapes, you can create an equation that represents the relationship between their dimensions. This is crucial for solving problems involving optimization where you might need to minimize or maximize an area or other function related to the shapes.

Minimum Total Area

The concept of minimizing total area arises in various optimization problems, where the objective is to find the dimensions that yield the smallest possible combined area of given shapes, under certain constraints. In our exercise, these constraints are defined by the sum of perimeters of the square and the equilateral triangle being constant.
When dealing with these problems, you'll need to express the area of each shape in terms of a common variable, which allows you to then derive a single equation that represents the total area. Applying the calculus concept of setting the derivative of this area function to zero helps you find the points where the area might be at a minimum or maximum. It's critical to combine the principles of geometry with those of calculus to reach the optimal solution.

Derivatives in Algebra

In algebra, derivatives are used to calculate the rate at which one quantity changes with respect to another. They are a powerful tool for finding the maxima and minima of functions, which is especially handy in optimization problems. A derivative is essentially the slope of the tangent to a function's graph at any given point and is represented by the function's prime notation (
for example,
\( f'(x) \)).
To find the minimum total area in our exercise, the derivative of the area with respect to one side of the shapes is set to zero. This action identifies the critical points, which might represent the minimum or maximum areas. After finding the critical points, a second derivative test can verify whether those points are minima, maxima, or points of inflection. If the second derivative is positive at the critical point, then the function has a local minimum there. The steps detailed in the provided solution showcase this process in action, helping students understand the practical application of derivatives in algebraic optimization problems.