Problem 6 - Thomas Calculus in SI Units

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 maximized when the line segment forms an isosceles triangle with equal legs, \\(a = b = \sqrt{200}\\).

Step 1: Understand the Triangle Setup

The line segment forms a right triangle in the first quadrant with the x-axis and y-axis as its legs. The coordinates of the ends of the line segment are \( (a, 0) \) and \( (0, b) \). The length of the hypotenuse is given as 20 units.

Step 2: Write the Equation for the Hypotenuse

According to the Pythagorean theorem, for the right triangle with legs \(a\) and \(b\) and hypotenuse 20, the equation is: \[ a^2 + b^2 = 20^2 \] Which simplifies to: \[ a^2 + b^2 = 400 \]

Step 3: Express Area in Terms of One Variable

The area \(A\) of the triangle is given by: \[ A = \frac{1}{2} \times a \times b \]Express \(b\) in terms of \(a\) using the equation from Step 2: \[ b = \sqrt{400 - a^2} \]Substitute \(b\) in the area formula: \[ A = \frac{1}{2} \times a \times \sqrt{400 - a^2} \]

Step 4: Differentiate Area with Respect to a

To find the maximum area, differentiate \(A\) with respect to \(a\) and find where it equals zero. Use the product rule: \[ \frac{dA}{da} = \frac{1}{2} \left( \sqrt{400 - a^2} + \frac{a \times (-\frac{a}{\sqrt{400 - a^2}})}{2} \right) \]Simplify using derivative rules.

Step 5: Solve Critical Point for Maximum Area

Solve \( \frac{dA}{da} = 0 \) to find critical points. Setting the differentiated equation to zero and simplifying gives:\[ \sqrt{400 - a^2} = \frac{a^2}{\sqrt{400 - a^2}} \]Cross-multiply and simplify: \[ 400 - a^2 = a^2 \]\[ 2a^2 = 400 \]\[ a^2 = 200 \]\[ a = \sqrt{200} \]

Step 6: Verify a=b Maximizes Area

Since \(a = b\), verify by substituting back into the original conditions. As previously \(a = \sqrt{200}\), calculate \(b = \sqrt{400 - a^2} = \sqrt{200} \). Both values are equal. Also, check the second derivative for concavity to confirm it's a maximum.

Key Concepts

Right Triangle

A right triangle is a type of triangle in which one of the angles is exactly 90 degrees, known as the right angle. This specific angle creates a distinct 3-sided shape where the two sides forming the right angle are referred to as the 'legs,' and the side opposite the right angle is the 'hypotenuse.' In the scenario where you're partitioning a section in a quadrant, the triangle assumes the x-axis and y-axis as its legs.

• The hypotenuse stretches between two points: one on the x-axis, (a, 0), and one on the y-axis, (0, b).
• The uniqueness of the right triangle lies in its predictable relationships, making it easier to handle with standard trigonometric approaches.

Understanding these basic elements offers clarity when applying different theorems or calculus methods, such as differentiation, to study its properties in depth.

Pythagorean Theorem

The Pythagorean Theorem is fundamental when studying right triangles. It establishes the relationship between the lengths of the sides of a right triangle. Specifically, it states that the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the legs (a and b). Mathematically, this is written as:\[ a^2 + b^2 = c^2 \]In optimization problems involving right triangles, such as finding the maximum area given certain constraints, the Pythagorean theorem plays a vital role. In our exercise scenario:

• We are given a hypotenuse of length 20 units.
• Thus, using the theorem, we form the equation \(a^2 + b^2 = 400\).
• This equation helps relate variables which are critical in applying calculus tools like differentiation.

By using the Pythagorean theorem, you can effectively determine other properties of the triangle needed to solve for optimization problems.

Differentiation

Differentiation is a process in calculus used to find how a function changes as its input changes. In the context of optimization, it helps to determine the maximum or minimum values of functions.For our task of maximizing the area of a right triangle:

• We described the area, \(A\), of the triangle as \(A = \frac{1}{2} \times a \times b\).
• To express in terms of one variable, \(b\) was replaced using \(b = \sqrt{400 - a^2}\) derived from the Pythagorean theorem.
• This substitution simplifies the function into a single-variable problem, \(A = \frac{1}{2} \times a \times \sqrt{400 - a^2}\).
• We then differentiate \(A\) with respect to \(a\) to find critical points where the area might be maximum.

Differentiation helps in analyzing these critical points by determining where the rate of change is zero, leading us to possible maximum values.

Critical Points

Critical points are a fundamental concept when it comes to optimization in calculus. They occur where the derivative of a function equals zero or is undefined, indicating potential points of maximum or minimum values.In this context:

• We set the derivative of the area function \(\frac{dA}{da}\) to zero to find the critical points.
• For the area function, solving \(\frac{dA}{da} = 0\) revealed that \(a^2 = 200\), thereby \(a = \sqrt{200}\).
• Substituting back the value of \(a\) verifies that \(b = \sqrt{200}\), confirming \(a = b\).
• Moreover, by analyzing the second derivative, we confirm that our critical point indeed provides a maximum area.

Thus, critical points are essential in not only finding where a function changes behavior but also in verifying the nature of these changes—whether they represent maximum or minimum values.