Question

A wire of length 340 mm is to be cut into two parts. One of the part is 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

Answer 1

The correct option is B 40

(b)
Let the length of two parts is x and y

$Sox+y=340mm..............\left(1\right)$

The part which has length x is used in making a square. So each side will be of length x/4, and the part which has length y is used in making a rectangle with sides in ratio 1 : 2. So sides have length (y/6 : y/3)

$Areaofsquareis\frac{x^2}{16}$

$Areaofrectangleis\frac{y^2}{18}$

So combined area is

$(\frac{x^2}{16}+\frac{y^2}{18})..............\left(2\right)$

Since the combined area should be minimum. So using equation 1 in 2, we get

$\frac{d}{dx}(\frac{x^2}{16}+\frac{{(340-x)}^2}{18})=0$

$[\frac{x}{8}+\frac{x}{9}-\frac{680}{18}]=0$

or x = 160 mm

and length of side of square is

$\frac{x}{4}=40mm$