Problem 9 - Calculus: Early Transcendentals

Question

Find the area of the largest rectangle that fits inside a semicircle of radius 10 (one side of the rectangle is along the diameter of the semicircle).

Step-by-Step Solution

Answer

The area of the largest rectangle is 100 square units.

Step 1: Understand the Problem

We need to find the maximum area for a rectangle that fits inside a semicircle with a radius of 10 units. One side of the rectangle will sit along the diameter of the semicircle.

Step 2: Model the Situation

The semicircle is on the coordinate plane with its diameter along the x-axis, centered at the origin. Its equation is \( y = \sqrt{100 - x^2} \). The rectangle fits inside this semicircle, so its width is \( 2x \) and its height is \( y \).

Step 3: Express the Area of the Rectangle

The area \( A \) of the rectangle is given by the formula \( A = \text{width} \times \text{height} = 2x \times y \). Substituting \( y = \sqrt{100 - x^2} \), we have \( A = 2x \sqrt{100 - x^2} \).

Step 4: Differentiate the Area Function

To maximize the area, we differentiate \( A(x) = 2x \sqrt{100 - x^2} \) with respect to \( x \), using the product rule and chain rule. We need to find \( \frac{dA}{dx} \) and set it to zero.

Step 5: Calculate the Derivative

The derivative is \( \frac{dA}{dx} = 2\sqrt{100-x^2} - \frac{2x^2}{\sqrt{100-x^2}} \). Simplify and solve for when \( \frac{dA}{dx} = 0 \) to find critical points.

Step 6: Solve for Critical Points

Set \( 2\sqrt{100-x^2} - \frac{2x^2}{\sqrt{100-x^2}} = 0 \), leading to \( 2(100-x^2) = 2x^2 \). Simplify this to obtain \( 100 = 2x^2 \). Solve to get \( x^2 = 50 \), hence \( x = \sqrt{50} = 5\sqrt{2} \).

Step 7: Verify Maximum Area

Substitute \( x = 5\sqrt{2} \) back into the area formula. Calculate \( y = \sqrt{100 - (5\sqrt{2})^2} = \sqrt{100 - 50} = \sqrt{50} \). So, area \( A = 2(5\sqrt{2})\sqrt{50} = 100 \).

Step 8: Conclusion

The area of the largest rectangle is 100 square units, at \( x = 5\sqrt{2} \).

Key Concepts

Maximizing Area of Rectangles

Optimizing the area of rectangles is a common problem in calculus, especially in geometric contexts like the one we have here. The goal is to find the maximum area that a geometric shape can accommodate under given constraints.

• In our scenario, the rectangle sits inside a semicircle with a fixed radius, which limits how large the rectangle can be.
• The diameter acts as one side of the rectangle, meaning the challenge is to maximize the area without crossing the boundaries of the semicircle.

The concept involves setting up an equation for the area of the rectangle in terms of variables, deriving this equation, and finding critical points where the maximum can occur.
This is a classic example of using mathematical tools to optimize situations, a crucial skill in calculus.

Semicircle Geometry

Semicircles are important geometric shapes, half of a full circle, and they help limit the space within which the rectangle here fits.

• Essentially, the semicircle in our problem has a radius of 10 units, centered at the origin on the x-axis.
• The equation of the semicircle is expressed as \( y = \sqrt{100 - x^2} \), which defines its upper boundary.

The rectangle must fit under this curve, with its bottom side lying along the diameter.
Understanding this geometry helps in creating the constraints that the rectangle's dimensions must adhere to. Using the geometry efficiently leads directly into calculus, where we can apply derivative techniques to compute solutions effectively.

Derivatives and Critical Points

Derivatives are the backbone of optimization in calculus, enabling us to find critical points which indicate potential maxima or minima of functions. In the context of maximizing the area of our rectangle, derivatives provide the mathematical method by which we determine the point at which the area is largest.

• First, we formulated the area function \( A = 2x \sqrt{100 - x^2} \) based on rectangle width and height.
• To find the maximum area, we differentiated \( A \), applying the product and chain rules to get \( \frac{dA}{dx} = 2\sqrt{100-x^2} - \frac{2x^2}{\sqrt{100-x^2}} \).

Setting the derivative to zero helps us find where the area stops increasing and begins to decrease, marking the point of maximum area.
Solving for these points and verifying with our conditions give us critical insights into the rectangle's optimal dimensions—demonstrating how calculus tools help solve geometric optimization challenges.