Problem 52 - Calculus Early Transcendentals

Question

Find the dimensions and area of the rectangle of maximum area that can be inscribed in the following figures. a. A right triangle with a given hypotenuse length \(L\) b. An equilateral triangle with a given side length \(L\) c. A right triangle with a given area \(A\) d. An arbitrary triangle with a given area \(A\) (The result applies to any triangle, but first consider triangles for which all the angles are less than or equal to \(90^{\circ}\).)

Step-by-Step Solution

Answer

A short answer based on the solution provided: For a right triangle with a given hypotenuse length \(L\), the dimensions of the maximum inscribed rectangle are \(\frac{L}{\sqrt{2}} \times \frac{L}{\sqrt{2}}\), and the maximum area is \(\frac{L^2}{2}\). For an equilateral triangle with a given side length \(L\), the dimensions of the maximum inscribed rectangle are \(\frac{L}{2} \times \frac{\sqrt{3}}{4} L\), and the maximum area is \(\frac{\sqrt{3}}{16}L^2\). For a right triangle with a given area \(A\), the dimensions of the maximum inscribed rectangle depend on its specific leg lengths, and an optimal solution can be achieved using a similar approach as discussed above by relating the area and leg lengths and then maximizing the inscribed rectangle's area using calculus.

Step 1: Set up the problem

Let's assume an inscribed rectangle with sides \(x\) and \(y\). Let the angle between the hypotenuse and the base of the right triangle be \(\theta\), then we can express \(x\) and \(y\) in terms of \(L\) and \(\theta\): \[x=L\cos{\theta}\] \[y=L\sin{\theta}\]

Step 2: Express the area of the rectangle

The area of the inscribed rectangle can be written as the product of the sides \(x\) and \(y\): \[\text{Area}=xy\]

Step 3: Substitute

Substitute \(x\) and \(y\) from Step 1 into the area equation: \[\text{Area}=(L\cos{\theta})(L\sin{\theta})\] \[\text{Area}=L^2\sin{\theta}\cos{\theta}\]

Step 4: Maximize the area

To maximize the area, we will find the derivative of the area with respect to \(\theta\) and set it equal to 0: \[\frac{d\text{Area}}{d\theta} = L^2 (\sin^2{\theta} - \cos^2{\theta})\] Setting \(\frac{d\text{Area}}{d\theta} = 0\), we have: \[\sin^2{\theta} = \cos^2{\theta}\] \[\tan^2{\theta} = 1\] \[\tan{\theta} = \pm 1\] Since \(0 \le \theta \le \frac{\pi}{2}\), \(\tan{\theta} = 1\) and \(\theta = \frac{\pi}{4}\).

Step 5: Find the dimensions and area

Substitute the value of \(\theta\) back into the equations for \(x\) and \(y\): \[x=L\cos{\frac{\pi}{4}}=\frac{L}{\sqrt{2}}\] \[y=L\sin{\frac{\pi}{4}}=\frac{L}{\sqrt{2}}\] The dimensions of the inscribed rectangle are \(\frac{L}{\sqrt{2}} \times \frac{L}{\sqrt{2}}\). The maximum area is: \[\text{Area}_{max}= \frac{L^2}{2}\] b. An equilateral triangle with a given side length \(L\)

Step 1: Set up the problem

Let's assume an inscribed rectangle with sides \(x\) and \(y\). Let the altitude of the equilateral triangle be \(h\). The altitude splits the equilateral triangle into two \(30\)-\(60\)-\(90\) right triangles, so we have: \[h = \frac{\sqrt{3}}{2} L\] Using similar triangles, the sides of the rectangle can be expressed in terms of \(h\) and \(y\): \[x = L - \frac{2y}{\sqrt{3}}\]

Step 2: Express the area of the rectangle

The area of the inscribed rectangle can be written as the product of the sides \(x\) and \(y\): \[\text{Area}=xy\]

Step 3: Substitute

Substitute \(x\) from Step 1 into the area equation: \[\text{Area}=(L - \frac{2y}{\sqrt{3}})y\]

Step 4: Maximize the area

To maximize the area, we will find the derivative of the area with respect to \(y\) and set it equal to 0: \[\frac{d\text{Area}}{dy} = L - \frac{4y}{\sqrt{3}} = 0\] Solve for \(y\): \[y = \frac{\sqrt{3}}{4}L\] Substitute \(y\) back into the equation for \(x\): \[x = L - \frac{2(\frac{\sqrt{3}}{4}L)}{\sqrt{3}} = \frac{L}{2}\] The dimensions of the inscribed rectangle are \(\frac{L}{2} \times \frac{\sqrt{3}}{4} L\). The maximum area is: \[\text{Area}_{max}= \frac{(\frac{L}{2})(\frac{\sqrt{3}}{4}L)}{2} = \frac{\sqrt{3}}{16}L^2\] c. A right triangle with a given area \(A\)

Step 1: Set up the problem

Let's assume an inscribed rectangle with sides \(x\)...

Key Concepts

Optimization in Calculus

Optimization is a fundamental concept in calculus that involves finding the maximum or minimum values of functions. In many real-world problems, we want to maximize or minimize quantities like cost, efficiency, area, volume, etc., using mathematical models. To solve such problems, we use derivatives – a central tool in calculus – which represent how a function changes as its inputs change.

For optimizing a function, we first need to express the quantity we want to optimize in terms of a variable. Then we find the function's derivative with respect to that variable and determine where the derivative is zero – these are our critical points. In the context of our exercise, we seek the dimensions of an inscribed rectangle that will give us the maximum area, and we use optimization techniques to solve it.

Maximum Area Problem

The maximum area problem is a classic optimization scenario where we want to find the dimensions that will give us the largest possible area. In geometric terms, this problem can involve shapes such as rectangles, triangles, or circles. The process involves setting up an equation for the area in terms of one or more variables, finding the derivative with respect to those variables to locate critical points, and then determining which of these provide a maximum area.

For example, with the inscribed rectangle in a right triangle example from our exercise, we used the properties of right triangles to express the rectangle's area in terms of the hypotenuse and an angle, and then we took the derivative to find the optimal angle that maximizes the area. Such problems exemplify the practical applications of calculus in geometry and various design fields.

Related Rates of Change

When we talk about 'related rates of change,' we are dealing with situations where two or more quantities are connected and change over time. Calculating related rates involves using derivatives to understand how the rate of change in one variable affects the rate of change in another. In optimization problems, we often don't deal with time directly, but the concept of related changes is still applicable as we look at how changes in one dimension of a shape can affect another dimension – especially when these dimensions are related by some constraint.

In the context of the inscribed rectangle, although time isn't a factor, we can still see the similarity to related rates as changes in the angle (or one side of the rectangle) affect changes in the overall area.

Derivatives Application

Derivatives are not just abstract concepts; they find practical application in many fields, including engineering, economics, physics, and computer science. In our optimization exercise, we applied the concept of derivatives by taking the derivative of the area function with respect to one of its variables (either an angle or a side). By setting this derivative equal to zero and solving, we found the maximum value of the area – a process that is applied in all kinds of optimization problems.

This application of derivatives is critical because it guides us to the optimal solution. It is by analyzing the derivative or the rate of change of a function that we make informed decisions – whether we're optimizing a function or making predictions based on data trends.