91Ó°ÊÓ

The distance formula is a crucial tool in geometry. It allows us to calculate the distance between two points in a coordinate plane. The formula is derived from the Pythagorean theorem and is expressed as:

• Distance: \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

In the problem, the line segment's endpoints are \((a, 0)\) and \((0, b)\), and the length is fixed at 20 units. Using the distance formula, we relate the two variables \(a\) and \(b\):

• \( \sqrt{a^2 + b^2} = 20 \)

This equation indicates that \(a\) and \(b\) cannot be arbitrarily chosen; they are dependent on one another while ensuring the segment remains 20 units long. Here, solving \(a^2 + b^2 = 400\) will help us explore the relationship between \(a\) and \(b\) to maximize the triangle's area.

The derivative is a fundamental concept in calculus, used to determine the rate at which a quantity changes. In optimization problems, like maximizing a triangle's area, derivatives help us find where changes occur and at what rate.
The area of the triangle, given by \( A = \frac{1}{2} ab \), depends on the values of \(a\) and \(b\). When one is expressed in terms of the other using the constraint \(b = \sqrt{400 - a^2}\), the area function becomes dependent on \(a\) only:

• \( A(a) = \frac{1}{2} a \sqrt{400-a^2} \)

Taking the derivative \( \frac{dA}{da} \) helps us find how the area changes with \(a\) and locate critical points where these changes yield maximum or minimum values. Here, applying rules like the product and chain rule, we arrive at the derivative:

• \( \frac{dA}{da} = \frac{1}{2} \left( \sqrt{400-a^2} - \frac{a^2}{\sqrt{400-a^2}} \right) \)

Using this derivative to find zero values tells us where the critical points are, key to maximizing the triangle's area.

Critical points are vital in calculus as they represent places where a function's rate of change is zero. This could mean a local maximum, minimum, or saddle point. To identify these points, we set the derivative equal to zero and solve.
In our problem, employing \( \frac{dA}{da} = 0 \) results in the equation:

• \( \sqrt{400-a^2} = \frac{a^2}{\sqrt{400-a^2}} \)

Upon simplifying, we find:

• \( 400 - a^2 = a^2 \)
• \( a^2 = 200 \)
• \( a = \sqrt{200} \)

Finding \(a\) allows us to pinpoint where the maximum area of the triangle occurs, with symmetry suggesting \(a = b\). Thus, these values not only satisfy the problem's constraints but also optimize the area.

The second derivative test is a method used to verify whether a critical point is a maximum or minimum. It provides insight into the function's concavity at the critical points.
Compute the second derivative \( \frac{d^2A}{da^2} \) of our area function. This step evaluates the concavity by revealing how the slope of \( \frac{dA}{da} \) changes past the critical point.

• If \( \frac{d^2A}{da^2} < 0 \): The function is concave down, confirming a local maximum.
• If \( \frac{d^2A}{da^2} > 0 \): The function is concave up, indicating a local minimum.

For this problem, the computation shows that \( \frac{d^2A}{da^2} \) is negative at \(a = b = \sqrt{200}\), verifying that this critical point indeed maximizes the triangle's area.
Hence, the triangle achieves its largest possible area when both \(a\) and \(b\) equal \(\sqrt{200}\).