problem maths?

Question

Please show me how to do this=

A length of wire , 4 metres long , is cut into two pieces.One piece is bent in the shape of a square, the other in a shape of a circle.Determine the radius of the circle so as to minimise the sum of the areas.

Answer 1

Cut the wire in half, you are left with 2 pieces 2 meters long.
One piece is bent in a square (2 meters /4 sides) = 0.5 meters per side. Area = 0.5 m * 0.5 m = 0.25 m^2

Second piece is formed into a circle. The circle's circumference is the length of the wire so,
Pi * D = 2 m or D = 2/Pi m so the radius R = D/2 = 1/Pi m

Area = Pi * R^2 = 1/Pi m^2 = 1/3.1416 m^2


There is no minimization required in this problem. When you form a square and a circle from fixed length items, the areas are fixed. The radius of the circle is fixed absolutely by the known circumference of the circle.

Answer 2

Only blitzmonster is correct so far.
With the length of the cut to form the square at 2.2404 meters, the area is a minimum at 0.5601m^2 and the radius of the circle is about 0.280049 m

Answer 3

The solution is best understood with the use of minimization.

So let x = length for the square part
y = 4-x = length for the circular part (since x+y=4m)

Note that radius is equal to (4-x)/(2*pi).

Sum of the areas = X^2+ pi*[(4-X)/(2*pi)]^2

Get the derivative of the expression and solve for x. Verify if x yields a minimum value and substitute to the expression for r to obtain the desired radius.

Cheers!

Answer 4

this can be solve by minima-maxima

let 4x be the perimeter of the square, x, the side of the square

(4 - 4x) be the circumference of the cirlcle, r = (4 - 4x)/(2)

take the derivative of sum of the areas then equate to zero

x^2 + (r^2) = Constant

x^2 + ((4 - 4x)/(2))^2 = Constant

simplifying this will give

(x^2( + 4) - 8x + 4)/ = Constant

taking the derivative, then equating to zero

2(x( + 4) - 4)/ = 0

then we can solve for x,

x = 4/( + 4)

and of course the radius

r = (4 - 4x)/(2)

substitute the value of x

r = (4 - 4·(4/( + 4)))/(2·)

r = 2/( + 4) or approx.

r = 0.2800495767

Answer 5

L = 4

4 s + 2 pi r = 4
=> s = (4- 2 pi r)/4 = (2- pi r)/2

Area = s^2 + pi r^2
A = [(2- pi r)/2]^2 + pi r^2
dA/dr = 2(2- pi r)/2*(- pi /2) + 2 pi r

solve dA/dr = 0 to find r

Answer 6

Too late to think... ill post answer tomorrow when my brain wakes up...