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### 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

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