Question

Find the dimensions of the rectangle of largest area that can

be inscribed in a circle of radius \(r\).

Short answer

The dimensions are \(2x = \sqrt 2 r\) and \(2y = \sqrt 2 r\).

Step-by-step solution

Application of Derivative

We can apply the derivative of a function to find the closest point of a curve to a point or the furthest point of a curve to another point. We find these points with the concept of local minimaand local maximarespectively.

We find local minima and maxima of a function by finding critical pointsand the second derivativeof the function.

Critical points are the \(x\) values that satisfy \(f'\left( x \right) = 0\).

The function has local maxima at a point \(x\) if \(f''\left( x \right) < 0\).

The function has a local minimum at a point \(x\) if \(f''\left( x \right) > 0\).

Finding equations related to a rectangle and cirlcle

Let the sides of the rectangle are \(2x,2y\).

Hence the area of the rectangle is \(4xy\).

Also, from the diagram we have \({r^2} = {x^2} + {y^2}\).

Hence \(y = \sqrt {{r^2} - {x^2}} \).

Now the area function will be

\(A\left( x \right) = 4x\sqrt {{r^2} - {x^2}} \).

Finding the Local Maxima of the area function

Differentiating we get,

\(A'\left( x \right) = 4\sqrt {{r^2} - {x^2}} - \frac{{4{x^2}}}{{\sqrt {{r^2} - {x^2}} }}\)

The critical point will be,

\(\begin{aligned}{c}A'\left( x \right) = 0\\4\sqrt {{r^2} - {x^2}} - \frac{{4{x^2}}}{{\sqrt {{r^2} - {x^2}} }} = 0\\4\left( {{r^2} - {x^2}} \right) - 4{x^2} = 0\\8{x^2} = 4{r^2}\\{x^2} = \frac{1}{2}{r^2}\\x = \frac{1}{{\sqrt 2 }}r\end{aligned}\)

Hence the critical point is \(x = \frac{1}{{\sqrt 2 }}r\).

The second derivative will be \(A'' = - 16x < 0\).

Hence the maximum value of the function \(A\) will be at the point \(x = \frac{1}{{\sqrt 2 }}r\).

Now,

\(\begin{aligned}{l}y = \sqrt {{r^2} - \frac{1}{2}{r^2}} \\y = \frac{1}{{\sqrt 2 }}r\end{aligned}\)

Hence the rectangle is a square.

Hence the dimensions are \(2x = \sqrt 2 r\) and \(2y = \sqrt 2 r\).