Question

A wire of length 34 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a rectangle whose length is twice its breadth. What should be the lengths of the two pieces, so that the combined area of the square and the rectangle is minimum?

Answer 1

Suppose the wire which is cut into two pieces of length x and y meters respectively.  
So, x + y = 34 m              .....(1)
Perimeter of square = 4(side) = x
Side = \[\frac{x}{4}\]

Area of square =  \[\left( \text { side } \right)^2 = \left( \frac{x}{4} \right)^2\]

Perimeter of rectangle = 2(l + b) = y

\[l + b = \frac{y}{2}\]

\[ \Rightarrow 2b + b = \frac{y}{2} (\text { Given } l = 2b)\]

\[ \Rightarrow b = \frac{y}{6}\]

Area of the rectangle = \[l \times b = 2b \times b = 2 b^2 = 2 \left( \frac{y}{6} \right)^2 = \frac{y^2}{18}\]

Now z = Area of square + area of rectangle

\[\Rightarrow z = \frac{x^2}{16} + \frac{y^2}{18}\]

\[ \Rightarrow z = \frac{x^2}{16} + \frac{\left( 34 - x \right)^2}{18} \left( \text { From } \left( 1 \right) \right)\]

\[ \Rightarrow \frac{dz}{dx} = \frac{2x}{16} + \frac{2\left( 34 - x \right)\left( - 1 \right)}{18}\]

\[ \Rightarrow \frac{dz}{dx} = \frac{x}{8} + \frac{\left( x - 34 \right)}{9} . . . . . (2)\]

For maximum and minimum values of z, \[\frac{dz}{dz} = 0\]

\[\frac{x}{8} + \frac{x - 34}{9} = 0\]

\[ \Rightarrow \frac{9x + 8x - 272}{72} = 0\]

\[ \Rightarrow 17x - 272 = 0\]

\[ \Rightarrow x = 16, y = 18\]

Now 

\[\frac{d^2 z}{d x^2} = \frac{1}{8} + \frac{1}{9} = \frac{17}{72} > 0\]

Hence, z is minimum when x = 16 and y = 18.
So, the wire should be cut into two pieces of length 16 m and 18 m.