We have a wire of length 28 meters and we have to form a circle and a square. What will be the maximum area and how? Can we draw the circle?
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There is only one circle and one square that you can form with a wire of 28m, if you use it all. For sure you have to use all of it. Can you find the radius of the circle and the size of the square ? Then compute the areas they enclosed and compare. Hint: what is illustrated here is called the isoperimetric inequality. – Tom-Tom May 20 '14 at 08:51
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Is the question which one of square or circle will have maximum area or you use the wire to form a square and a circle. If the question is the former, Circle will have the maximum area than square for a given perimeter. – Satish Ramanathan May 20 '14 at 08:54
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Do you have to make both the square and the circle, and decide how much wire to allot to each one? – user2357112 May 20 '14 at 09:30
2 Answers
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The circle will have the circumference $C=28 $, meaning that its radius $ r=\frac {C}{2\pi}=\frac {14}{\pi} $, which in turn means that its area $ A_c =\pi r^2=\frac {196}{\pi} $. The perimeter of the square will be $P=28 $, meaning that its side $ a=\frac P4=7 $, which in turn means that its area $ A_s=a^2=49 $. Comparing $ A_c$ and $ A_s $ we get $A_c> A_s$.
user132181
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I mean practically is it possible to have a circle of radius 14/(pi)....like squaring a circle! you know what I mean – I.R May 20 '14 at 09:13
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Mathematically, it is possible. Squaring the circle is a whole different problem. If by 'practically' you mean 'in the real world', then I'd say no. – user132181 May 20 '14 at 09:16
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@IdeedRafiqi We cannot draw any circle (of any radius) in "the real world". We can approximate it, though. – naslundx May 20 '14 at 09:19
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Let $x$ be the perimeter of the square, and $y$ the perimeter of the circle. So you have $x+y=28$. The area of the square would be $\frac{x^2}{16}$ (why?). The area of the circle would be $\frac{y^2}{4\pi}$ (why?). You did not specify exactly what you want to maximize, but I guess it is the sum of areas: $$\frac{x^2}{16}+\frac{y^2}{4\pi}.$$
Do you know how to take it from here?
Added: I see no response from the OP, but I'll finish the answer myself in case anyone is interested in this. Plugging $y=28-x$ into the above equation you get $$\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}=\frac{x^2}{16}+\frac{784-56x+x^2}{4\pi}= \left(\frac{1}{16}+\frac{1}{4\pi}\right)x^2-\frac{56}{4\pi}x+\frac{784}{4\pi}.$$ Now, you want to find a maximum for the above function for $x\in [0,28]$. Note that since the coefficient of $x^2$ is positive, the function attains its maximum at $x=0$ or $x=28$, so substituting $x=0$ you get $\frac{784}{4\pi}=\frac{196}{\pi}=62.388737\dots$, while substituting $x=28$ you get $\frac{784}{16}+\frac{784}{4\pi}-\frac{784}{2\pi}+\frac{784}{4\pi}=49$.
That is, the answer is that to get the greatest area you should use the whole wire to make a circle.
Note that the answer by @user132181 just computes the area of a circle with perimeter $28$, the area of a square with perimeter $28$ and compares them. He has not shown that this is indeed the maximum sum of areas.
DKal
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