Question

A piece of wire 12 m long is cut into two pieces. One piece is bent into the shape of a circle of radius r and the other is bent into a square of side s. How should the wire be cut so that the total area enclosed is:

a) a maximum r= and s= ?

b) a minimum r= and s= ?

Answer 1

hero got it

Answer 2

Thanks for having so much confidence in me

Answer 3

no calc for this one, pure reason gets it i think

Answer 4

think square

Answer 5

s=\[\pi r ^{2}+(12-r)^{2}=\S\] for the max u should ds/dr =0 so u could find r

Answer 6

sri u can't use 12-r
as jero said 2pi r + s=12 , u should find r from this equation

Answer 7

sri hero ! i make a mistake when i typed u name

Answer 8

Don't pay any attention to the equations I wrote

Answer 9

Actually, that equation is wrong

Answer 10

maximum, make a circle.
minimum, make a square

Answer 11

for a given length of wire \(l\), if it is bent into a circle we get:\[2\pi r=l\quad\implies \pi r=\frac{l}{2}\quad\implies Area=\pi r^2=\frac{l^2}{4}\]and if bent into a square with side length \(s\) we get:\[4s=l\quad\implies s=\frac{l}{4}\quad\implies Area=s^2=\frac{l^2}{16}\]so, as @satellite73 said, maximum area is achieved for a circle, minimum for a square.

Answer 12

You should explain the minimum and maximum part

Answer 13

sorry - mistake in area of circle, should be:\[2\pi r=l\quad\implies r=\frac{l}{2\pi}\quad\implies Area=\pi r^2=\frac{l^2}{4\pi}\]

Answer 14

and:\[\frac{l^2}{4\pi}>\frac{l^2}{16}\]since \(\pi\) is less than 4.

Answer 15

You kind of short-cutted the explanation by using l instead of 12 - s

Answer 16

I was trying to explain that, for a given length of wire, the circle will enclose a bigger area than a square.

Answer 17

The length for the circle is different from the length for the square, but yet you use the same variable for both without distinguishing between the two

Answer 18

for a given length of wire, if we bend it into a circle we will get a larger area than if we bend it into a square.
so the next steps would be to cut the wire of length 12 into two such that one piece is of length x and the other is of length 12-x and work out what gives us a maximum/minimum enclosed area.
what I showed was just the first step to establish that a circle always encloses a larger area than a square for a given length of wire.

Answer 19

therefore maximum is achieved when entire wire is used to create a circle where:\[2\pi r=12\quad\implies r=\frac{6}{\pi}\text{ and }s=0\]and minimum area is achieved when the entire wire is used to create a square where:\[4s=12\quad\implies s=3\text{ and }r=0\]

Answer 20

You kinda skipped a whole bunch of steps there, but okay

Answer 21

I entered the minimum but it was wrong...

Answer 22

What minimum did you enter?

Answer 23

s=0 r =3?

Answer 24

wait i mean the other way..

Answer 25

s= 3 and r=0...

Answer 26

for a) it refers to a maximum and we know the circle has the maximum therefore, so s = 12 r = 0
for b) it refers to a minimum and we know that the square has a minimum s = 0, but the square cannot be defined in terms of r because it doesn't have a radius.

Answer 27

ok guys - looks like this problem was a bit more subtle than I thought.

the problem states find the maximum and minimum "total area enclosed".
so if asume we cut the 12m wire into two with one wire of length \(x\) used for the square and the remaining wire of length \(12-x\) used for the circle, then we get:\[2\pi r=12-x\quad\implies r=\frac{12-x}{2\pi}\quad\implies A_{circle}=\pi r^2=\frac{(12-x)^2}{4\pi}\]and:\[4s=x\quad\implies s=\frac{x}{4}\quad\implies A_{square}=s^2=\frac{x^2}{16}\]so the total area A is given by:\[A=A_{circle}+A_{square}=\frac{(12-x)^2}{4\pi}+\frac{x^2}{16}\]to get the minimum/maximum we need to set the differential of A to zero, so:\[A'=\frac{-2(12-x)}{4\pi}+\frac{2x}{16}=\frac{x}{8}-\frac{12-x}{2\pi}=0\]this gives:\[\frac{x}{8}=\frac{12-x}{2\pi}\]\[2\pi x=96-8x\]\[\pi x=48-4x\]\[(4+\pi)x=48\]\[x=\frac{48}{4+\pi}\]the second derivative of A gives us:\[A''=\frac{1}{8}+\frac{1}{2\pi}>0\]therefore we know \(x=\frac{48}{4+\pi}\) MUST represent a minimum value.

we therefore get the following for the minimum:\[x=\frac{48}{4+\pi}\]\[r=\frac{12-x}{2\pi}=\frac{12-\frac{48}{4+\pi}}{2\pi}=\frac{\frac{48+12\pi-48}{4+\pi}}{2\pi}=\frac{12\pi}{2\pi(4+\pi)}=\frac{6}{4+\pi}\]\[s=\frac{x}{4}=\frac{12}{4+\pi}\]

the maximum remains at using the entire wire for the circle so \(r=\frac{6}{\pi}\), \(s=0\).

Answer 28

thanks! ^-^

Answer 29

yw