It’s given that the radius of the circle is 1. \(r=1\)
Let’s assume that h is the height of rectangle and w is the width of rectangle.
The area of the rectangle is defined by the formula below:
\(A=w*h\)
We can calculate the diagonal of the rectangle by using Pythagorean theorem
\(d^2 = w^2 + h^2\)
Let’s assume that the largest rectangle has diagonal that equals to circle diameter.
\(diameter=2*r=2*1=2\)
\(d^2 = w^2 + h^2 = diameter^2=2^2\)
\(w^2 + h^2 = 4\)
\(w = \sqrt{4-h^2}\)
\(A=h*\sqrt{4-h^2}\)
To find the critical points, we take the derivative of A and set it equal to 0, then solve for x.
h <- Sym("h")
A <- h*sqrt(4 - h^2)
PrettyForm(Simplify(deriv(A,h)))##
## 2 2
## 4 - h - h
## --------------
## / 2 \
## Sqrt\ 4 - h /
##
## <OMOBJ>
## <OMS cd="logic1" name="true"/>
## </OMOBJ>Derivative equals to \(A'=\frac{4-2h^2}{\sqrt{4h^2}}\)
Let’s solve for \(A'=\frac{4-2h^2}{\sqrt{4-h^2}}=0\)
\(\sqrt{4-h^2}\) must be greater than 0. So that, \(4-h^2\) must be grater than 0. Consequently, x falls into the interval (-2;2).
\(4-2*h^2=0\)
\(h^2=2\)
\(h_1=\sqrt2\) and \(h_2=-\sqrt2\) (looks unrealistic since height is a positive value)
Lets find width \(w = \sqrt{4-\sqrt{2}^2}=\sqrt{4-2}=\sqrt2\)
The dimensions of the rectangle with largest area that can be drawn inside the unit circle are
height equals to \(\sqrt2\)
width equals to \(\sqrt2\)