Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Minimizing an Enclosed Area


Date: 06/02/99 at 12:39:39
From: Scott Miller
Subject: Calculus

Consider the quadrant I portion of the graph of the curve x = y^2 from 
the point Q(4,2) where the curve intersects the line y = 2, to the 
origin O(0,0). Let P(x1,2) be a variable point in quadrant I on the 
line y = 2 where 0 < x1 < 4. Determine the position of P so that the 
line segments OP and PQ and arc OQ of the given curve enclose minimum 
total area.

So far, I've drawn this and I've tried using the equations y = mx+b 
and (y-y1) = m(x-x1) to try to figure out the equation of line OP. I 
really have no clue how to start this or where to go with it. I know 
I'm trying to minimize area and I think I need to use integration but 
I can't figure out how to use it.


Date: 06/06/99 at 01:41:52
From: Doctor Pete
Subject: Re: Calculus

Hi,

The first thing to observe is that OP intersects OQ. So, the area you 
are looking at is actually two distinct regions. The second thing you 
might notice is that you can reflect the entire problem about the line 
y = x, so that the problem is to find the area enclosed by

     y = x^2,    (this is OQ)
     x = 2,      (PQ)
     y = mx,     (OP)

where m is a real number between 0 and 2. Since OP intersects OQ, 
there is an interval where x^2 > mx, and there is an interval where 
mx > x^2. The point of intersection, of course, is where x^2 = mx, 
that is, x = m. So we notice that on the interval [0,m], the area 
enclosed is

     Integral[mx - x^2, {dx, 0, m}].

And on the interval [m,2], x^2 > mx, hence the area enclosed is

     Integral[x^2 - mx, {dx, m, 2}].

The first integral is simply

     m^3/6,

and the second is

     8/3 - 2m + m^3/6.

So the total area enclosed is

     A(m) = m^3/3 - 2m + 8/3,

again where 0 <= m <= 2. This area is minimal when A'(m) = 0; i.e.,

     A'(m) = m^2 - 2 = 0,

or m = Sqrt[2]. It follows that P has the coordinate (2, 2*Sqrt[2]). 
But remember we reflected the coordinate axes to make the problem 
easier, so the answer is actually

     P = (2*Sqrt[2],2).

Finally, observe that the two separate integrals can be written as a 
single integral:

     A(m) = Integral[|x^2-mx|, {dx, 0, 2}],

where |x| is the absolute value of x.  

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/