Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius 10 inches?

Answer 1

Common sense would suggest that the optimal rectangle would be a square, but here is a rigorous proof. Consider one of the rectangle's sides to be of length s. The rectangle's diagonal is a chord. Since we know that it subtends an angle of 90 degrees at the circumference, we know that it is also a diameter. Its length is therefore always 20 inches. Given this much, we can use Pythagoras' theorem to deduce the length of the other side, t. t = sqrt(202 - s2) = sqrt(400 - s2) = (400 - s2)0.5 The area of the rectangle, A, is the product of its two sides. A = st = s * (400 - s2)0.5To find the maximum value of A, we need to differentiate this function of s. Consider the function to be a product of two functions u and v, where u represents s and v denotes (400 - s2)0.5. A' and u' and v' represent the differentials of A and u and v with respect to s. So from the product rule of differentiation: A' = uv' + vu' The first problem is to find v'. This requires the chain rule. v' = 0.5 x (400 - s2)-0.5x -2s = -s(400 - s2)-0.5 And v' is of course just 1. So we have: A' = -s2(400 - s2)-0.5 + (400 - s2)0.5 It turns out that this simplifies to A' = (400 - 2ss) / sqrt(400 - x2) The maximum (and minimum) values of A are found where A' = 0. For this to happen, it is evident that (400 - 2ss) must equal zero. 400 - 2ss = 0 2s2 = 400 s2 = 200 soptimal = sqrt(200) Substituting this into our expression for the other side, t, we find that t is also sqrt(200) hence, as expected, we have a square. The square is, of course, of area 200 square inches. sqrt(200), incidentally, is 10sqrt(2) or about 14.142 inches. It is simple enough to generalise all the working, and produce this result: * The largest rectangle that can be inscribed in a circle of radius R is a square of side-length Rsqrt(2) and area 2R2.