A rectangle is inscribed in a circle of radius 8m

Question

A rectangle is inscribed in a circle of radius 8m

a. Find the dimensions of the rectangle that will maximize the area of the rectangle

b. Find the maximum value of the area

Answer 1

Since this looks like homework, I'm going to set up the problem and leave the calculus to you. Leave a comment if you get stuck.

After rotating, we can assume that the sides are horizontal/vertical. Then notice that if we know the coordinates of the vertex in the first quadrant, we know the coordinates of all 4 vertices. Using the usual notion of angle as a parameter, the vertex in the first quadrant has coordinates $(8\cos(\theta), 8\sin(\theta))$, where $0\leq\theta\leq\frac{\pi}{2}$. The area of the part of the rectangle in the first quadrant (which is a quarter of the whole rectangle) is therefore $8\cos(\theta)\cdot8\sin(\theta)$.

Answer 2

Here's another approach.

If the rectangle is ABCD with points in order then AC is a diagional of the rectangle and a diameter of the circle (angle in a semicircle). Call the centre of the circle O.

The area of the rectangle can be split into two parts ABC and CDA which are congruent.

Consider the diameter AC as the base of triangle ABC. To maximise the area of the triangle one wants to maximise the height, since the base is fixed. By drawing a line through B parallel to AC it is easy to see that the height is maximum when OB is perpendicular to AC.

It is then straightforward to do the basic calculations to get the answer. No calculus.

Answer 3

Of course, we can assume the circle's centered at the origin of coordinates and etc.

Call the rectangle's sides' lengths $\,x\,,\,y\,$ , then $\,x^2+y^2=8^2=64\,$ (why? An inscribed angle's measure is $\,90^\circ\,$ iff the cord which it subtends...).

From here, you want to minimize/maximize the function $\,f(x,y)=xy\,$ subject to the above condition, which is the same as $\,y=\sqrt{64-x^2}\,$ on the upper semicircle of the circle . Well, can you continue from here?

Added Let us take the circle $\,x^2+y^2=(8m)^2=64m^2\,$ , and let $\,R\,$ be a rectangle inscribed in the circle with sides parallel to the axis and thus with vertices $$(x,y)\,,\,(x,-y)\,,\,(-x,-y)\,,\,(-x,y)$$

From here, its perpendicular sides' lengths are $\,2x\,,\,2y\,$ (horizontal and vertical, resp.).

By Pythagoras, and since both diagonals of $\,R\,$ cross through the origin, we get $$(2x)^2+(2y)^2=(16m)^2\Longrightarrow 4(x^2+y^2)=256m^2\Longrightarrow x^2+y^2=64m^2$$

We thus get that we must maximize the function $\,f(x,y)=(2x)(2y)\,$ subject to the condition $\,y=\sqrt {64m^2-x^2}\,$ (the circle's upper semicircle. Note that all is symmetric here wrt. the origin and axis), so our function to maximize becomes

$$f(x)=x\sqrt{64m^2-x^2}\Longrightarrow f'(x)=\sqrt{64m^2-x^2}-\frac{x^2}{\sqrt{64m^2-x^2}}=0\Longleftrightarrow $$ $$\Longleftrightarrow 64m^2-2x^2=0\Longrightarrow x^2=32m^2\Longrightarrow x=4\sqrt 2\,m\Longrightarrow $$ $$\Longrightarrow y=\sqrt{64m^2-32m^2}=4\sqrt 2\,m$$ And indeed we get a square with sides $\,2x=2y= 8\sqrt 2\,m\,$ and the maximal area is $\,(8\sqrt 2\,m)^2=128m^2\,$