Dimensions of Rectangle

Question

Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r=9.

What are the dimensions of the rectangle?

Answer 1

Let us inscribe in a circle of radius 9 a rectangle ABCD.

Obviously the point of intersection of the diagonals AC and BD

is the center of the circle.

Triangle ABC is a right triangle at B whose hypotenuse AC is 9 units.

Let us call ∠ ACB = θ

then AB = 18 sinθ and BC = 18 cosθ

Hence the area of the rectangle ABCD is A(θ) = (18 sinθ)(18cosθ)= 162 sinθ2

A^'( θ )= 324cos2θ

A^'( θ ) = 0 implies cos2θ = 0 that is 2θ =90 and finally θ= 45º critical point

Then A^''( θ ) = -81sin2θ and A^''( 45)= -81 <0.

Then by the second derivative test the function A(θ) = (162) sinθ2 has a max at θ=45.

This implies that the rectangle inscribed in a circle with maximum area is a square.

In our case the side of the square has length 9√2.

The area of the inscribed square is A = ( 9√2)² =162

Answer 2

the largest area is a square

9 is the hypotenuse of a right triangle with sides = 9/sqr2

sides of the square are 2 times 9/sqr2 = 9sqr2 = about 9(1.414) = 12.728

area of the square = (9sqr2)^2 = 81(2) = 162