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Question

A wire 80 cm long is cut into 3 pieces of lengths y, x and 80 − x − y and each piece is bent into squares. Find the absolute maximum and absolute minimum of the areas of these squares. Please help by showing me how you did it?

Answer 1

The absolute maximum is for x=y=0 and is A= 400cm^2
Sum A = x^2/16 +y^2/16 +z^2/16
x+y+z -80=0
Lagrangian multipliers
Extremes of
F(x,y,z) = x^2+y^2+z^2 with x+y+z-80=0

H(x,y,z,k)= x^2+y^2+z^2 +k(x+y+z-80)
Hx = 2x+k=0
Hy= 2y+k=0
Hz= 2z+k=0
x=y=z =-k/2
k(-3/2k-80)=0 k= 0 not a solution and
k= -160/3 so x= 80/3cm
and Sum A = 1/16 (6400/3)=133.33 cm^2
(minimum)

Answer 2

wire 1: length = y, so each side of the square = (1/4)y
area = (1/4)y*(1/4)y = (y^2)/16

wire 2: length = x, so each side = (1/4)x
area = (1/4)x*(1/4)x = (x^2)/16

wire 3: length = 80 - x - y, so each side = 20 - (1/4)x - (1/4)y
area = (20 - (1/4)x - (1/4)y)^2
area = (x^2)/16 + (xy)/8 - 10x + (y^2)/16 - 10y + 400

so, those are the three possible areas made from the three pieces of wire.

The absolute max. area would be if x and y = 0, so you only have one square from an 80 cm long wire. the area of that would be 300 cm^2.

The absolute min. area would be 0 cm^2

Answer 3

These squares would have to be equal all around (sides are equivalent).
y/4 is one of the four sides ... area = y^2/16
y^2/16 + x^2/16 + (80 - x - y)^2/16
y^2/16 + x^2/16 + (80 - x - y)(80 - x - y)/16
y^2/16 + x^2/16 + (6400 - 160x - 160y + x^2 + 2xy + y^2)/16
y^2/16 + x^2/16 + 6400/16 - 160x/16 - 160y/16 + x^2/16 + 2xy/16 + y^2/16
y^2/8 + x^2/8 + 400 - 10x - 10y + xy/8
From there, you have your expression of the area...

Answer 4

Wow, thats a really pukey problem, what math are you in???