/sci/ - Science & Math

http://www.wolframalpha.com/
http://www.khanacademy.org/

Because the area of a triangle can be demonstrated as half of the area of a paralellogram which has the same base length and height, and the maximum area of the rectangle in question is achieved when the corners touch two of the triangle's medians.

Go figure.

if you take pi times pi plus the circumference of a square and divide it by the number of times it takes plutonium to half life you have ur answer!!

Willing to help, but I don't know the answer, so I drawed the triangle to visualize...

>To help you visualise it, vertice P of the rectangle would touch AB, vertice Q would touch AC, and vertices R and S both touch BC. Good luck.
If you can prove that fact (it doesn't follow trivially from the fact that the rectangle has maximal area), you can try to express the area of the rectangle as a function of its height, and set the derivative to zero.

let point b be at the origin (0,0), and call the slope of BA m, the slope of AC n (a negative number), and the length of BC L.
the equation of the line that BA is a segment of is y=mx, the equation of the line that AC is a segment of is y=n(x-L).
These two lines intersect when mx = n(x-L), so x=nL/(n - m), so they would intersect at (some x, mnL/(n-m)). this means the area of the triangle A = mnL^2/(n-m)
Next, let the length of BR be called the variable k. We can determine the height of the rectangle (PR = QS) to be km. so we can find the length of SC to be km/n (a negative number). since we know the length of BC = L. then RS = L - k + km/n
so the area of the rectangle B = km(L - k + km/n) = kmL - k^2m + k^2m^2/n
if we differentiate with respect to k we get. dB/dk = mL - 2km + 2km^2, which we want equal to 0 to find the maximums and minimums.
0 = mL - 2km + 2km^2, k = -L/(2m-2)

Fuck, i totally made a mistake somewhere didnt I.
Do your own fucking homework.

>>1959710
I lol'd.

>>1959710
>cant figure it out
>do your own hw
you should probably do it for your own sake

>>1959726
Don't worry, I just figured out a much easier way to do it geometrically

you only have to prove it for a right triangle

dropping a perpendicular from base to apex will make two right triangles

easy to show half is attainable by rectangle half height of triangle

prove it's maximal with a bit of easy calculus i guess

>>1959813
>prove a square is maximal area rectangle you can fit in a right triangle
PROTIP: it isn't.

>>1959826
Smells like bullshit. The square is clearly a stationary point in the area coverage.

>>1959841
My counterexample disagrees.

>>1959813
A square will only be the maximum area obtainable if you're talking about perimeter. In this case, it's not.

I'm saying this because was my 1st try, too :-(

>>1959850
Oh. Miscommunication.
I meant the rectangle with a vertex that bisects the hypotenuse. That's always right. For the special case of a right triangle with equal-length legs, that rectangle is a square.

>>1959865
That's probably correct, though I'm still not sure on how to prove it. >>1959631 (samefag) only works if you can prove that the largest rectangle has the right angle as one of its vertices, which isn't trivial at all.

>>1959883
>only works if you can prove that the largest rectangle has the right angle as one of its vertices, which isn't trivial at all.
The only case I think you'd have to confront is a rectangle that has an edge along the hypotenuse, instead of edges along the legs.

>>1959899
That rectangle happens to have the same size, by the way.
So how do you prove that the rectangle must have one of its edges along an edge of the triangle? Again, it appears obvious, but proving it ain't trivial.

See pic. With the same reason, we deduct n = c/2.
Now, it's easy to prove that the area in the rectangle is equal the area outside the rectangle.