Expert: Azeem Hussain - 3/9/2009

Question
the area of the yard is 100 sq m.  To put a fence around the yard the
cheapest way at a cost of $20/m for the length and $5/meter for the width is
the question.  What is the length and width.
I know that if the length is x and the width is y, then xy=100 and I know that
2x+2y=50 (maybe).  But that's as far as I go.  Can you help?  

Answer
Hi Rob!

Let x equal length and y equal width.  xy=100.  (Your assumption about the perimeter is not applicable.)  If xy=100, then y=100/x.
Now make an expression for the cost.  Let's call this C, and x is the variable.
C(x)=20x+5y
C(x)=20x+500/x

You want to find this function's minimum.  Differentiate.
C'(x)=20-500/x^2

You want to find the critical points, as they may tell you the local minima and maxima.  Critical points may occur where the function is not differentiable, or where C'(x) equals 0.

0=20-500/x^2
20=500/x^2
x^2=500/20
x^2=25
x=5

(x could also equal -5, but looking at the real life situation, x must be in the interval ]0,100[.)

The length is 5 and the width is 100/5=20.

Thanks for asking,
Azeem