Problem 6 - Thomas Calculus

Question

You are planning to close off a corner of the first quadrant with a line segment 20 units long running from \((a, 0)\) to \((0, b) .\) Show that the area of the triangle enclosed by the segment is largest when \(a=b.\)

Step-by-Step Solution

Answer

The area is largest when \(a = b = \sqrt{200}\).

Step 1: Expressing the Triangle's Vertices

The triangle's vertices are given as \((a, 0)\), \((0, b)\), and the origin \((0,0)\). These define a right triangle in the first quadrant.

Step 2: Calculating the Length of the Line Segment

The length of the line segment from \((a,0)\) to \((0,b)\) is given by the distance formula: \(\sqrt{a^2 + b^2} = 20\). This is the constraint we will use.

Step 3: Solving the Constraint for One Variable

We solve for \(b\) in terms of \(a\) using the equation \(\sqrt{a^2 + b^2} = 20\). Squaring both sides, we have \(a^2 + b^2 = 400\). Therefore, \(b = \sqrt{400 - a^2}\).

Step 4: Expressing the Area of the Triangle

The area of the right triangle is given by \(\text{Area} = \frac{1}{2} \times a \times b\). With \(b\) expressed in terms of \(a\), the area becomes \(\text{Area} = \frac{1}{2} a \cdot \sqrt{400 - a^2}\).

Step 5: Setting Up the Function to Maximize

We aim to maximize the area function \(A(a) = \frac{1}{2} a \sqrt{400 - a^2}\). This requires finding the derivative \(A'(a)\) and solving \(A'(a) = 0\).

Step 6: Calculating the Derivative

We use the product rule to find \(A'(a)\): \(A(a) = \frac{1}{2} \cdot a \cdot (400 - a^2)^{1/2}\). The derivative is computed to be \(A'(a) = \frac{1}{2} \left[ (400 - a^2)^{1/2} + a \cdot \left( \frac{-a}{\sqrt{400 - a^2}} \right) \right]\), simplifying this gives \(A'(a) = \frac{1}{2} \left( \frac{400 - 2a^2}{\sqrt{400 - a^2}} \right)\).

Step 7: Finding the Critical Points

To find the critical points, solve \(\frac{400 - 2a^2}{\sqrt{400 - a^2}} = 0\), leading to \(400 = 2a^2\) which simplifies to \(a^2 = 200\) and thus \(a = \sqrt{200}\).

Step 8: Verifying the Maximum

At \(a = \sqrt{200}\), since \(a=b\), we check if this leads to the largest area. \(b = \sqrt{400 - a^2} = \sqrt{200}\), hence the area is \(\frac{1}{2} \times \sqrt{200} \times \sqrt{200} = 100\). This is consistent with maximizing the area formula for the given constraints.

Key Concepts

Triangle Area Maximization

In calculus, optimization involves finding the maximum or minimum values of a function under given constraints, and here, we are dealing with maximizing the area of a triangle within specific boundary conditions. The triangle in question has vertices at

• \((a, 0)\)
• \((0, b)\)
• The origin \((0,0)\)

This forms a right triangle. By maximizing the area of this triangle, we improve the composition of the sides to adhere to the maximum value allowed by the constraint. Ultimately, our goal is to show that when

• \(a = b\)

The area is at its largest possible value. This forms the basis of the problem's optimization challenge.

Distance Formula

The distance formula is a fundamental tool in geometry and calculus. It helps us express the length between two points in a plane. For two points,

• \((x_1, y_1)\)
• \((x_2, y_2)\)

The distance \(d\) is calculated as:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In this exercise, the line segment from

• \((a,0)\)
• \((0,b)\)

must have a fixed length of 20 units. When we apply the distance formula within our context:\[\sqrt{a^2 + b^2} = 20\]This equation acts as a constraint for optimizing the triangle's area.

Derivative Calculation

Calculating the derivative is essential for finding the maximum or minimum of a function, particularly in problems involving optimization. The derivative helps us understand how a function's output changes as its input changes. For our problem, we set up a function to express the area of the triangle:\[A(a) = \frac{1}{2}a\sqrt{400 - a^2}\]To find the derivative \(A'(a)\), we apply the product rule, which is used when a function is the product of two or more functions. The product rule states:\[(fg)' = f'g + fg'\]With this rule, differentiating the area function provides insight into how the area changes as we vary \(a\). Solving for critical points involves setting this derivative to zero, guiding us towards finding the maximum area configuration.

Critical Points Analysis

Critical points are specific values in the domain of a function where the function's derivative is zero or undefined. These points are vital in determining local maxima or minima, which are essential in optimization tasks. To locate these points for our triangle area function:\[A'(a) = \frac{1}{2} \left( \frac{400 - 2a^2}{\sqrt{400 - a^2}} \right) = 0\]Solving this equation yields a crucial part of the solution:

• \(a^2 = 200\)
• Therefore, \(a = \sqrt{200}\)

At this point, the derivative switches from positive to negative, indicating a maximum rather than a minimum. Verifying that \(a = b = \sqrt{200}\) gives the maximum area confirmed by substituting back into the constraints and the area function:\[\text{Area} = \frac{1}{2} \times \sqrt{200} \times \sqrt{200} = 100\]This confirmation reassures us that our critical point gives the largest possible area for the triangle given the constraints.