q (2 Viewers)

[Arithela]

A piece of wire 6m long is cut into 2 parts, one of which is used to form a square and the other to form a rectangle whose length is 3 times its width. Find the lengths of the two parts if the sum of the areas is a minimum.

I got the lengths awfully close to the correct answers, so is anyone willing to provide worked solutions so i can see where i went wrong in the working out?

 

[undalay]

A piece of wire 6m long is cut into 2 parts, one of which is used to form a square and the other to form a rectangle whose length is 3 times its width. Find the lengths of the two parts if the sum of the areas is a minimum.

I got the lengths awfully close to the correct answers, so is anyone willing to provide worked solutions so i can see where i went wrong in the working out?

Let x be the piece that forms the square
Let each side be of length x/4
Let y by the piece that forms the rectange.
Let longer sides be of length 3y/8 and shorter side y/8

x+y = 6
x=6-y
Thus length side of square is 6-y/4

A = (6-y)^2 + (y/8)(3y/8)
A = y^2 -12 y +36 + 3y^2/64
A = 67y^2/64 - 12y + 36
A' = 67y/32 - 12 = 0
67y = 384
y = 384/67
x = 18/67

Is that right? I might of made a careless mistake somewhere.

edit: u need to prove its a minimum don't forget. y"=0

 

[ssglain]

Let x be the piece that forms the square
Let each side be of length x/4
Let y by the piece that forms the rectange.
Let longer sides be of length 3y/8 and shorter side y/8

x+y = 6
x=6-y
Thus length side of square is 6-y/4

A = (6-y)^2 + (y/8)(3y/8)
A = y^2 -12 y +36 + 3y^2/64
A = 67y^2/64 - 12y + 36
A' = 67y/32 - 12 = 0
67y = 384
y = 384/67
x = 18/67

Is that right? I might of made a careless mistake somewhere.

edit: u need to prove its a minimum don't forget. y"=0

That should have been [(6 - y)/4]²

I've attached the solution.

 

[Arithela]

ahh... my first step was

let x be a side of square and y = side of rectangle

therefore

4x + 8y = 6


thanks guys