divide region into 2 equal areas

[galactus]

Here is an interesting little problem from another site(I got permission to post it). I thought it would be fun to see the different ways it's tackled, if anyone has a mind to.

"In the first quadrant, a horizontal line y = c cuts the curve \(\displaystyle y = 8x - 27x^3\) into two regions under the curve. Find the constant c such that the areas of the two regions are equal".

Because the line intersects in 2 places and is horizontal, it makes it a little more challenging, I believe. Also, it's difficult to solve for y, which would make it easier.

 

[galactus]

Well, no one bit. I thought this was a fun problem. Oh well, here's my approach if anyone is interested. The approach I got from the person who showed me this problem is a little different and a little less messy. They used the top part of the graph above the line, but the answer is the same.


It is best to use some sort of technology for this problem, lest you have some mighty ornery algebra to contend with.


If we draw a horizontal line, y=c, through the graph of \(\displaystyle 8x-27x^{3}\), we see it intersects in 2 places. Call these (a,c) and (b,c)

It is easily seen that the entire region is area 16/27. Therefore, we are looking for a line which divides the region into two subregions of area 8/27.



So we have:

\(\displaystyle \L\\c=8a-27a^{3}\)

\(\displaystyle \L\\c=8b-27b^{3}\)

Setting equal and solving for b(using technology, of course), we get:

\(\displaystyle \L\\b=\pm\frac{(\sqrt{-3(81a^{2}-32)}-9a)}{18}\)

Now, we can define our humongous functions with respect to 'a' only.

Using the bottom part of the graph below the line:

\(\displaystyle \L\\\int_{0}^{a}(8x-27x^{3})dx+\int_{b}^{\frac{2\sqrt{6}}{9}}(8x-27x^{3})dx+c(b-a)=\frac{8}{27}\)

Solving this monstrosity(with Maple's help) leads to a solution of a=.0779462908

Subbing back into \(\displaystyle 8a-27a^{3}\), we get

c=.6107838722

 

[pka]

Here is a historical Post Script.
I think that I recall first seeing this problem in a preliminary edition of the calculus by Gleason, Hughes-Hallett, et al in the mid-eighties. (This calculus series came to be known as the ‘Harvard Calculus’ reformed text.) As I remember it, the problem was dropped after the first or second edition. It is just a nightmare of algebra and got in the way of the calculus.

I see you gave the image back.
Could it be the *.jpeg ?

 

[galactus]

I believe Ted must've done something. I was just able to post the graph a few minutes ago.

Yes, the problem has some Herculean algebra, but I liked it. Thank heavens for CPU's.