A 25 inch copper wire will be used to form two squares. This is done by cutting the 25 inch
wire into two parts. Minimize the possible sum of the two areas of the square
2 answers
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Expecting the two squares to be equal .... anyway
let side of the first square be x
let the side of the 2nd square be y
4x + 4y = 25
y = (25-4x)/4
total area = x^2 + y^2
= x^2 + ((25 - 4x)/4)^2
= x^2 + (625 - 200x + 16x^2 )/16
d(total area)/dx = 2x - 25/2 + 2x
= 4x - 25/2
= 0 for a min of area
4x = 25/2
x = 25/8
y = (25 - 4(25/8)) / 4 = 25/8
sure enough, since x = y, the two squares are equal
with each square equal to (25/8)^2 square units
let side of the first square be x
let the side of the 2nd square be y
4x + 4y = 25
y = (25-4x)/4
total area = x^2 + y^2
= x^2 + ((25 - 4x)/4)^2
= x^2 + (625 - 200x + 16x^2 )/16
d(total area)/dx = 2x - 25/2 + 2x
= 4x - 25/2
= 0 for a min of area
4x = 25/2
x = 25/8
y = (25 - 4(25/8)) / 4 = 25/8
sure enough, since x = y, the two squares are equal
with each square equal to (25/8)^2 square units
6 answers