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Maximum area of a rectangle inscribed in a circle

Use the circle equation to remove one variable

For a centered rectangle with first-quadrant vertex $(x, y)$ on a radius- $r$ circle, width $= 2 x$ , height $= 2 y$ , and $x^{2} + y^{2} = r^{2}$ . Therefore $A = 4 x y = 4 x r^{2} - x^{2}$ . The area is greatest when $x = y = r / 2$ , so the rectangle is a square.

Problem

A rectangle centered at the origin is inscribed in the circle $x^{2} + y^{2} = 36$ . Its first-quadrant vertex is $P (x, y)$ . Express its area in terms of $x$ and find the maximum area.

Answer

A (x) = 4 x 36 - x^{2}

; maximum area

72

square units at

x = 32 \approx 4.24

The maximum is exact; 4.24 is only a rounded coordinate.

Step-by-step solution

1. Build the area function

The width is $2 x$ and height is $2 y$ . Since $y = 36 - x^{2}$ in the first quadrant,

A (x) = 4 x y = 4 x 36 - x^{2}, 0 < x < 6.

2. Maximize without rounding

Because $A$ is nonnegative, maximize its square.

A (x)^{2} = 16 x^{2} (36 - x^{2}) .

Put $u = x^{2}$ . Then the variable part is $36 u - u^{2}$ , a downward-opening parabola with vertex at $u = 18$ . Hence $x = 18 = 32$ and $y = 32$ .

3. Find the area

A_{m a x} = 4 (32) (32) = 72.

The rectangle is a square with side $62$ .

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