Problem 37 - Essential Calculus

Question

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

To find the dimensions of the equilateral triangle and the square that yield a minimal total area, follow the steps outlined above. This involves setting up the equations with the given conditions, finding the derivative, and ensuring that the second derivative of the Area function is positive at these points.

Step 1: Set Up Equations

Let the side length of the square be \(x\) and the side length of the equilateral triangle be \(y\). Then the perimeter of the square is \(4x\) and the perimeter of the triangle is \(3y\). Given that the sum of the perimeters is 10, we can express this as: \(4x + 3y = 10\).\n\nNext, let's establish formulas for the area of each shape. Using the side lengths, we know that the area of the square is \(x^2\) and the area of an equilateral triangle is \(\frac{\sqrt{3}}{4}y^2\). So the total area \(A\) expressed in terms of \(x\) and \(y\) is: \( A = x^2 + \frac{\sqrt{3}}{4}y^2\).

Step 2: Express Total Area as a Single Variable Function

From the perimeter equation (\(4x + 3y = 10\)), we may isolate \(y\): \(y = \frac{10 - 4x}{3}\). Substituting this into the Area equation we get: \(A = x^2 + \frac{\sqrt{3}}{4}\left(\frac{10 - 4x}{3}\right)^2\).

Step 3: Find the Derivative of the Area Function

To optimize the Area function (minimize the area), its critical points need to be found. The derivative of the Area function with respect to \(x\) will yield a function that equals zero at the critical points. Differentiate \(A\) with respect to \(x\): \(A' = 2x - \frac{4\sqrt{3}(10-4x)}{9}\).

Step 4: Find the Critical Points

Set the derivative equal to zero and solve for \(x\): \(2x - \frac{4\sqrt{3}(10-4x)}{9} = 0\). Solving this equation will yield the value of \(x\).

Step 5: Confirm the Minimum

Once the critical points have been calculated, plug these values back into the Area function to calculate the corresponding values of \(y\). These are the side lengths of the square and the triangle. To confirm that the calculated dimensions give a minimum total area and not a maximum, ensure that the second derivative of the Area function is positive. If it's positive, your results correspond to a global minimum.