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About Azeem Hussain
Expertise
I can answer virtually any kind of question dealing with of Math 536 and below, my forte being in parabolic functions and analytical geometry. I'm currently learning Linear Algebra, and cannot answer questions that deal with subject matter more advanced than that.

Experience
I am neither a professor nor a teacher of this subject. I am merely a student who is gifted at mathematics and enjoys being of service to his community. I frequently tutor people in math and the results are usually great.

Publications
Reflections, Riverside School Board (2005, 2006)

Education/Credentials
Diploma of Secondary Studies from Chambly Academy High School, and IBO-MYP certificate as well. My lowest mark on a high school math final was 97%, peaking at 99% in 2006 and 2007 (second-highest Math 436 mark in the province). Being a Quebecer, I am fluent in English and French and can respond to questions easily in both languages.

Awards and Honors
Pascal Math Competition, School Champion(2007)

 
   

You are here:  Experts > Science > Math for Kids > Geometry > Calculus I - Minimizing Cost

Geometry - Calculus I - Minimizing Cost


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

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