A wire of length 36 mm is to be cut into two parts. One of the parts is to be made into a square and the other into a rectangle where sides are in the ratio of 1:2. What is the length of the side of the square (in mm) such that the combined area of the square and the rectangle is a MINIMUM? A. 30 B. 40 C. 120 D. 180

GATE 2016 · General Aptitude · Geometry · medium

Answer: Side of square = 2L/17 = 40 mm (with wire length L = 340 mm)

1. Set up the constraint: Let the wire length be L. Square perimeter = 4a, rectangle perimeter = 6w. So 4a + 6w = L, which gives w = (L - 4a)/6.
2. Express total area as function of a: A(a) = a^2 + 2*(L-4a)^2/36 = a^2 + (L-4a)^2/18
3. Find minimum by differentiation: dA/da = (68a - 8L)/18 = 0. So 68a = 8L, a = 8L/68 = 2L/17. For a = 40: L = 40*17/2 = 340 mm. Check: 4*40 + 6w = 340 => 160 + 6w = 340 => 6w = 180 => w = 30. Area = 1600 + 2*900 = 3400.
4. Verify it is a minimum: The second derivative is positive, confirming a = 40 mm gives a minimum (not maximum).