question giving me guff

[momomath]

In triangle ABC, angle B = 90 degrees and AC has a constant length. Prove that the area of triangle ABC is a maximum when AB = BC.

I labeled AB "A", BC "B", and AC "C".

Since it is a right triangle, I started with A^2 +B^2 = C^2. Then i rewrote it as A = √(C^2 - B^2).

The area of a triangle is 1/2(base)(height), so i figured i would input A for the base, and B for the height:

Area = (√(C^2 - B^2))(B)/2

Now i've hit a wall though

 

[momomath]

Do not use capitals for lengths: AB = a, BC = b (not necessary to deal with AC).

Assume you're dealing with a rectangle size a by b.
A right triangle is exactly half a rectangle...OK?

Prove are is maximum when a = b (a square is a special rectangle).
Go here for example:
http://answers.yahoo.com/question/index?qid=20090121162314AA5oVd5

I'm sorry, I should have mentioned that I'm in a quadratics section of my book. So I have to prove it by creating a quadratic equation. So I think I have to find how 'a' is a function of 'b', and how the area is a function of that first function. i have to prove it by getting this information into the vertex form of a quadratic equation.

 

[Deleted member 4993]

In triangle ABC, angle B = 90 degrees and AC has a constant length. Prove that the area of triangle ABC is a maximum when AB = BC.

I labeled AB "A", BC "B", and AC "C".

Since it is a right triangle, I started with A^2 +B^2 = C^2. Then i rewrote it as A = √(C^2 - B^2).

The area of a triangle is 1/2(base)(height), so i figured i would input A for the base, and B for the height:

Area = (√(C^2 - B^2))(B)/2

Now i've hit a wall though

Let

AB = m and

AC = k (constant length Hypotenuse)

Then

BC² = k² - m²

Area = 1/2 * AB * BC → Area² = 1/4 * m² * (k² - m²)

Again let

Area² = y

and

x = m²

then we have

y = 1/4 * x * (k² - x) ← there is your quadratic equation and continue....

 

[momomath]

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thank you so much! i was forgetting some of my exponent rules when i did it in my notebook.