Problem 5 - Calculus for Biology and Medicine

Question

Find the largest possible area of a right triangle whose hypotenuse is \(4 \mathrm{~cm}\) long.

Step-by-Step Solution

Answer

The largest possible area is 4 cm².

Step 1: Understand the Problem

We need to find the triangle with the greatest area. The triangle in question is a right triangle, and its hypotenuse is given as 4 cm. The area of a triangle is calculated by the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). For a right triangle, the base and height are the two legs of the triangle.

Step 2: Apply the Pythagorean Theorem

For a right triangle with hypotenuse \( c = 4 \), and legs \( a \) and \( b \), the Pythagorean theorem states \( a^2 + b^2 = c^2 \). Since \( c = 4 \), we have \( a^2 + b^2 = 16 \).

Step 3: Express One Variable in Terms of the Other

Express \( b \) in terms of \( a \): \( b = \sqrt{16 - a^2} \). This helps us express the area in terms of one variable.

Step 4: Write the Area as a Function of a Single Variable

The area \( A \) is given by \( A = \frac{1}{2} \times a \times b = \frac{1}{2} \times a \times \sqrt{16 - a^2} \).

Step 5: Maximize the Area Function

To find the maximum area, set the derivative of \( A \) with respect to \( a \) to zero. Calculate \( \frac{dA}{da} = \frac{1}{2} \left( \sqrt{16 - a^2} - \frac{a^2}{\sqrt{16 - a^2}} \right) \). Simplify the expression to solve \( \frac{dA}{da} = 0 \).

Step 6: Solve for Critical Points

After simplification, solving \( \sqrt{16 - a^2} = \frac{a^2}{\sqrt{16 - a^2}} \) gives \( a^2 = 8 \), so \( a = 2\sqrt{2} \). Substituting \( a = 2\sqrt{2} \) into \( b = \sqrt{16 - a^2} \) gives \( b = 2\sqrt{2} \).

Step 7: Calculate the Maximum Area

Substitute \( a = 2\sqrt{2} \) and \( b = 2\sqrt{2} \) into the area formula: \( A = \frac{1}{2} \times 2\sqrt{2} \times 2\sqrt{2} = \frac{1}{2} \times 8 = 4 \).

Step 8: Verify Solution

Verify using the Pythagorean theorem: \((2\sqrt{2})^2 + (2\sqrt{2})^2 = 8 + 8 = 16\). The maximum area is consistent with our calculations.

Key Concepts

Understanding the Hypotenuse in Right Triangles

A hypotenuse is the side opposite the right angle in a right triangle. It is always the longest side of the triangle. In any right triangle, understanding the hypotenuse is critical, as it plays a central role in calculations involving sides and angles. In our case, the hypotenuse of the right triangle is given as 4 cm.

In a right triangle, the other two sides are called legs. These legs are adjacent to the right angle, and they work together with the hypotenuse to describe the triangle's geometry fully. The length of the hypotenuse offers vital information that allows us to dig deeper into the triangle's properties using various mathematical tools and theorems. One such tool is the Pythagorean theorem, which we'll discuss next.

The Pythagorean Theorem

The Pythagorean theorem is an essential mathematical principle used with right triangles. It states: *In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.* The theorem is expressed with the formula:

• \( a^2 + b^2 = c^2 \)

Where \( a \) and \( b \) are the legs and \( c \) is the hypotenuse. For our triangle, we substitute the given hypotenuse value to get \( 4^2 = 16 \). This allows us to set up the equation \( a^2 + b^2 = 16 \), which connects the legs of our triangle to the hypotenuse.

The Pythagorean theorem not only helps us understand the relationships between the sides of a triangle but also enables us to solve for unknown side lengths when any two sides are known. In our problem, it plays a crucial role in helping us express one variable in terms of the other, facilitating the maximization of the area.

Maximizing the Area of a Right Triangle

To maximize the area of a right triangle with a specific hypotenuse, we focus on the area formula for triangles:

• \( ext{Area} = \frac{1}{2} imes ext{base} imes ext{height} \)

For right triangles, the base and height correspond to the legs. Therefore, maximizing the area means finding the values of the legs that satisfy the condition for the largest possible area.

In our case, with a hypotenuse of 4 cm, we use the Pythagorean theorem to express one leg in terms of the other. By stating \( b = \sqrt{16 - a^2} \), we rewrite the area formula as a function of a single variable: \( A = \frac{1}{2} \times a \times \sqrt{16 - a^2} \). To find the maximum area, we need to calculate the derivative, set it to zero, and solve for critical points. Doing the math, we find that when both legs are \( 2\sqrt{2} \) cm, the area reaches its maximum value of 4 square centimeters. This calculative approach demonstrates how mathematical principles can optimize geometric properties.