Problem 25 - Calculus Early Transcendentals

Question

A rectangle is constructed with its base on the diameter of a semicircle with radius 5 and its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

Step-by-Step Solution

Answer

Answer: The dimensions of the rectangle with maximum area are x = √50 and y = √12.5.

Step 1: Setting Up Variables and Relationships

Let's denote the length of the rectangle as x and the width as y. Since the base of the rectangle is on the diameter, we can say x is the diameter of the semicircle, so x = 2 * radius = 10. We'll create a right-angled triangle with sides x (length), y (width), and h (distance from the vertex on the semicircle to the diameter), as shown below: ``` _ | ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅| y ├──────|──────┤ x/2 x/2 ``` Since the rectangle is symmetric, the distance from the vertex on the semicircle to the midpoint of the diameter is x/2. Now, we'll use the Pythagorean theorem to find the relationship between x, y, and h.

Step 2: Pythagorean theorem

We can apply Pythagorean theorem on the right-angled triangle: \((x/2)^2 + y^2 = h^2\) Since the radius of the semicircle is 5, h = 5. Substituting that into the equation: \((x/2)^2 + y^2 = 5^2\) Now we want to express y in terms of x to create an equation that represents the area of the rectangle in terms of a single variable.

Step 3: Express y in terms of x

We can solve the equation for y: \(y^2 = 5^2 - (x/2)^2\) \(y = \sqrt{25 - (\frac{x^2}{4})}\) The area of the rectangle can be found by multiplying its length and width: \(A(x) = x \cdot y\) \(A(x) = x \cdot \sqrt{25 - (\frac{x^2}{4})}\)

Step 4: Find the Maximum Area Using Calculus

To find the maximum area, we need to differentiate A(x) with respect to x and set the derivative equal to 0. First, let's find the derivative of A(x): \(A'(x) = \frac{d}{dx} \left( x \cdot \sqrt{25 - (\frac{x^2}{4})} \right)\) We can apply the product rule here: \((uv)' = u'v + uv'\) Let u = x and v = \(\sqrt{25 - (\frac{x^2}{4})}\), then \(u' = 1\) \(v' = \frac{d}{dx} \sqrt{25 - (\frac{x^2}{4})} = \frac{-x}{4\sqrt{25 - (\frac{x^2}{4})}}\) Now applying the product rule to find the derivative of A(x): \(A'(x) = 1 \cdot \sqrt{25 - (\frac{x^2}{4})} + x \cdot \frac{-x}{4\sqrt{25 - (\frac{x^2}{4})}}\) Now, set the derivative equal to 0 to find the critical points: \(0 = \sqrt{25 - (\frac{x^2}{4})} - \frac{x^2}{4\sqrt{25 - (\frac{x^2}{4})}}\) We can multiply both sides of the equation by \(4\sqrt{25 - (\frac{x^2}{4})}\) to get rid of the square root: \(0 = 4(25 - (\frac{x^2}{4})) - x^2\) Solve for x: \(0 = 100 - x^2 - x^2\) \(x^2 = 50\) \(x = \sqrt{50}\)

Step 5: Find the dimensions of the rectangle

Now that we have the value of x, we can find y: \(y = \sqrt{25 - (\frac{x^2}{4})}\) \(y = \sqrt{25 - (\frac{50}{4})}\) \(y = \sqrt{25 - 12.5}\) \(y = \sqrt{12.5}\) So, the dimensions of the rectangle with the maximum area are x = \(\sqrt{50}\) and y = \(\sqrt{12.5}\).