Exercise 4.3, Problem 17 (pg. 187)

What are the dimensions of the rectangle with the largest area that can be drawn inside the unit circle?

Solution

We know that the unit circle has a radius of 1 and a diameter of 2. We will use this fact to draw out our problem.

We are trying to maximize the dimensions of the rectangle, so we will take the derivative of the rectangle’s area \((A=xy)\) and set equal to \(0\) to do so. In order to write \(y\) in terms of \(x\), we can use the fact that the sides of our rectangle and the radius form a right triangle and rearrange this formula. In other words: \[2^2 = x^2 + y^2\] \[ y^2 = 4 - x^2\] \[ y = \sqrt{4 - x^2}\]

Let’s substitute this into our area formula: \[A = xy\] \[A(x) = x(\sqrt{4 - x^2})\]

We will now take the derivative of the area using the product rule: \[A'(x) = (\sqrt{4 - x^2}) + \frac{x * -2x}{2\sqrt{4-x^2}}\] \[A'(x) = (\sqrt{4 - x^2}) - \frac{x^2}{\sqrt{4-x^2}}\]

To maximize, we will set this derivative to \(0\) and solve: \[0 = (\sqrt{4 - x^2}) - \frac{x^2}{\sqrt{4-x^2}}\] \[\frac{x^2}{\sqrt{4-x^2}} = (\sqrt{4 - x^2})\] \[x^2 = 4 - x^2\] \[2x^2 = 4\] \[x^2 = 2\] \[x = \sqrt{2}\]

We can now solve for \(y\): \[ y = \sqrt{4 - x^2}\]
\[ y = \sqrt{4 - \sqrt{2}^2}\] \[ y = \sqrt{4 - 2}\] \[ y = \sqrt{2}\]

We can see from this that \(x=y\), which means the largest rectangle is actually a square!