# A rectangle is inscribed in a circle of radius 8m [closed]

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

-

## closed as off-topic by Jonas Meyer, JimmyK4542, Macavity, Behaviour, Arthur Fischer♦Sep 15 at 4:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, JimmyK4542, Macavity, Behaviour, Arthur Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.

What have you tried/thought about so far? Also, this looks like homework. That's fine, but you should use the appropriate tag if it is. –  Brett Frankel Aug 14 '12 at 18:23
It seems that you are trying to find an answer of 4m x 4m which is on an answer sheet. If you try drawing anything like an accurate scale diagram you will see that a 4m x 4m square falls far short of the maximum if the diameter of the circle is 16m as you have given. –  Mark Bennet Aug 14 '12 at 20:20
-1 It makes sense that the optimum is a square, as demonstrated in the answers. If you suspect that the answer sheet is wrong, you could check if it fits the circle. In this case it does not. Please show your work (the 4m by 4m only came out in comments) as it makes it much easier to give a useful answer. –  Ross Millikan Aug 15 '12 at 3:23

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 $8\cos(\theta)\cdot8\sin(\theta)$.

-
Um. No. The dimensions are supposed to be 4m by 4m but I can't get the answer. I do not know where trigonometry came from.... –  user37600 Aug 14 '12 at 18:33
@user37600 "Um. No." is not a particularly polite way to ask for help, especially when Brett's setup is completely correct. The trigonometry came in when he was trying to find the coordinates of the upper-right corner of the rectangle (draw a picture). Using these coordinates, the total area of the rectangle (in terms of $\theta$) is $4 \cdot 8\cos(\theta) \cdot 8\sin(\theta)$. It remains to maximize this function using calculus. –  Austin Mohr Aug 14 '12 at 18:39
I didn't say it impolitely. I was just confused about how trigonometry could solve this problem. Actually, this problem does not involve trigonometry at all. I don't even understand how this problem could be solved with trigonometry. It has to include Pythagorean theorem. –  user37600 Aug 14 '12 at 18:50
@user37600: Further to Brett Frankel's excellent approach, note that $64\sin\theta\cos\theta=32\sin 2\theta$. But $\sin 2\theta$ has maximum value $1$, when $\theta=\pi/4$. So the maximum area of the quarter rectangle is $32$. Look, Ma, no calculus! –  André Nicolas Aug 14 '12 at 18:54
@user37600: In my earlier comment (erased since it may cause confusion) I was referring to the first quadrant part. The whole thing is $8\sqrt{2}$ by $8\sqrt{2}$. –  André Nicolas Aug 14 '12 at 19:03

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.

-
Nice. I was going to say no calculus is needed, but you beat me to it. –  Michael Hardy Aug 14 '12 at 20:27

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 $\,(8\sqrt 2\,m)^2=128m^2\,$

-
I did that but I didn't get the answer 4m by 4m. The answer sheet could be wrong –  user37600 Aug 14 '12 at 18:52
The answer in your sheet is 4m by 4m?? And the data you gave is accurate? Then there is no doubt: that sheet is wrong, and in fact it is wrong big time. –  DonAntonio Aug 15 '12 at 3:28