Question

Need Help! Will give medal!!!
A piece of wire 4 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

For maximum,
dA/dx=0
let parameter of square is x
So area=x^2/16
Then 2πr=(4-x)
r=4-x/2π
Area=πr^2=(4-x)^2/4π
Add area of square and circle and differentiate it to 0 ,then find x

Answer 2

still a little lost

Answer 3

I have no idea what you are talking about

Answer 4

fine :p

do you want to walk through it? ie a solution?

Answer 5

In which part

Answer 6

both

Answer 7

I think you are wanting a completecomplete solution

Answer 8

@kdancer03
@Pawanyadav split the wire into lengths x and y

then used the x length to make a square and the y length to make a circle

for the square, if it has perimeter x, it has area \(\dfrac{x^2}{16}\) ie \(\dfrac{x}{4} \times \dfrac{x}{4}\)

good so far?

Answer 9

that I get

Answer 10

the remaining bit of length y forms a circle of circumference y. and from \(C = 2 \pi r\) we know that that circle's radius can be gotten from \(y = 2 \pi r\)

so \(r = \dfrac{y}{2 \pi}\)

its area is \(A = \pi r^2 = \pi (\dfrac{y}{2 \pi})^2 = \dfrac{y^2}{4\pi}\)

ok?

Answer 11

yes

Answer 12

and because x + y = 4, we have x = 4-y and y = 4-x

so we can convert the square area to a term in y, or the circle are to a term in x

and then combine them

make sense? or should we just do it?

Answer 13

recap
\(A_{s}(x) = \dfrac{x^2}{16}\)

\(A_{c}(y) = \dfrac{y^2}{4\pi}\)

and

\(x + y = 4\)

Answer 14

I gotcha still

Answer 15

so now you decide

do you go for A(y) or A(x), reducing this to single variable calculus. and then you set \(A' = 0\) to find a stationary point, ie max/min.....