Expert: Azeem Hussain - 3/25/2009

Question
Hi Azeem..
OK OK this is the original problem:
a 30 cm wire is cut to two. One piece is to bend to a square. The other is to a rectangular with ratio of 2:1 length-to-width. Question: What are the lengths of the two pieces if sun of the areas of thesquare and rectangular is a minimum?

Thank you much Azeem..
erica

Answer
Hi Erica!

Let the perimeter of the rectangle equal x and that of the square equal y.  x+y=30, so y=30-x.

Given that rectangle has a length to width ratio of 2:1, that means its long side will measure one third of the perimeter and the shorter side will be one sixth of the perimeter (such that the sum of all the sides is x).  The rectangle's area is (x/3)(x/6)=x²/18.

The square's perimeter will be y/4.  Its area will be y²/16, or more appropriately, (30-x)²/16.

The function for area will be the sum of both areas, so A=x²/18+(30-x)²/16.  This can be simplified as A=17x²/144+15x/4+225/4.  This is the function that needs to be minimized.  This can be done by using calculus, finding the derivative, finding its critical point...

Alternatively, find the axis of symmetry of the function [x=-B/(2A), or find the zeroes and take half the distance between them].  x turns out to be approximately 15.88.  y=30-x, so 14.12.

SOLUTION: The square is made out of 14.12 cm of wire and the rectangle out of 15.88 cm wire.

Thanks for asking,
Azeem