# Thread: Maximum area of a triangle

1. ## Maximum area of a triangle

Given a right angle triangle with integral lengths for all 3 sides. Let a, b ,c be the sides of the triangle, and $\displaystyle a<b<c$ and a+c=81. Find the maximum area of this triangle under these conditions.

I can't think of a way to do this without using calculus. I got 47 as the length of b that will give the largest area.

2. Originally Posted by chengbin
I can't think of a way to do this without using calculus. I got 47 as the length of b that will give the largest area.
Since c is the largest side, it is the hypotenuse.

Therefore

$\displaystyle A=\frac{1}{2}b\cdot{a}$

You know that $\displaystyle c=\sqrt{a^2+b^2}$ so $\displaystyle a+\sqrt{a^2+b^2}=81$Solving for a

$\displaystyle (\sqrt{a^2+b^2})^2=(81-a)^2$
$\displaystyle a^2+b^2=81^2-162a+a^2$

$\displaystyle b^2=81^2-162a$

$\displaystyle a=\frac{81^2-b^2}{162}$

Now substitute and maximize

3. Originally Posted by VonNemo19
Since c is the largest side, it is the hypotenuse.

Therefore

$\displaystyle A=\frac{1}{2}b\cdot{a}$

You know that $\displaystyle c=\sqrt{a^2+b^2}$ so $\displaystyle a+\sqrt{a^2+b^2}=81$Solving for a

$\displaystyle (\sqrt{a^2+b^2})^2=(81-a)^2$
$\displaystyle a^2+b^2=81^2-162a+a^2$

$\displaystyle b^2=81^2-162a$

$\displaystyle a=\frac{81^2-b^2}{162}$

Now substitute and maximize
That's exactly the way I did it, and I got 47. But this requires calculus. I need a way to do this problem without using calculus.

BTW, can/would manipulating the values of a and c change the area?

4. Originally Posted by chengbin
That's exactly the way I did it, and I got 47. But this requires calculus. I need a way to do this problem without using calculus.

BTW, can/would manipulating the values of a and c change the area?
I thought you said you wanted to do it WITH calculus. Sorry. I must have read your post wrong.

5. Originally Posted by VonNemo19
I thought you said you wanted to do it WITH calculus. Sorry. I must have read your post wrong.
Any ideas?

6. Originally Posted by chengbin
Any ideas?
Intuitively I Know that a right triangle will have it's largest area when a=b. Maybe there's something there.....

Maybe still do the substitution but only evaluate at the endpoints of the feasable domain. I'm sure that'll work.

7. Originally Posted by VonNemo19
Maybe still do the substitution but only evaluate at the endpoints of the feasable domain. I'm sure that'll work.
We don't know the endpoints of the domain.

Anyway, this was on my friend's worksheet from this tutor center. I seriously doubt they taught calculus, that's why I have to exclude the use of calculus, even though that this solution is my first instinct when I look at that question.

8. Hello, chengbin!

Edit: I corrected my typos . . . Thanks, yeongil!

This is more of a Discrete Math problem, not Calculus.
I did some experimenting and came up with a solution, sort of.

Given a right angle triangle with integral lengths for all 3 sides.
Let $\displaystyle a, b ,c$ be the sides of the triangle, and $\displaystyle a<b<c$ and $\displaystyle a+c=81.$
Find the maximum area of this triangle under these conditions.
We want three positive integers $\displaystyle a,b,c$ where: .$\displaystyle \begin{array}{cc}{\color{blue}(a)} & a<b<c \\ {\color{blue}(b)} & a^2+b^2 \:=\:c^2 \\ {\color{blue}(c)} & a+c \:=\:81 \end{array}$

Here's the experimental part . . .

From (b) we have: .$\displaystyle b^2 \:=\:c^2 - a^2 \:=\:(c+a)(c-a)$

Sustituting (c), we have: .$\displaystyle b^2 \:=\:81(c-a)$
. . and we see that $\displaystyle (c-a)$ must be a square.

Since $\displaystyle a+c \:=\:81$, it narrows down the choices.

I found four solutions: .$\displaystyle (a,b,c) \:=\:\begin{array}{cc}(1) & (40,9,41) \\ (2) & (36,27,45) \\ (3) & (28,45,53) \\ (4) & (16,63,65) \end{array}$

Solutions (1) and (2) violate condition (a),
. . which leaves us with solutions (3) and (4).

. . $\displaystyle \begin{array}{|c|c|}\hline (a,b,c) & \text{Area} \\ \hline \\[-5mm] (28,45,53) & {\color{red}630} \\ (16,63,65) & 504 \\ \hline\end{array}$

Hope I didn't miss any . . .
.

9. @Soroban

How did you get those numerical values? Guess?

I don't understand the part where you say a-c must be a square.

10. Originally Posted by chengbin
I don't understand the part where you say a-c must be a square.
Earlier, Soroban said:
From (b) we have: .$\displaystyle b^2 \:=\:a^2 - c^2 \:=\a+c)(a-c)$

Sustituting (c), we have: .$\displaystyle b^2 \:=\:81(a-c)$
. . and we see that $\displaystyle (a-c)$ must be a square.
First, a minor correction: it should be
$\displaystyle b^2 \:=\:{\color{red}c}^2 - {\color{red}a}^2 \:=\c + a)(c - a)$
and
$\displaystyle b^2 \:=\:81({\color{red}c - a})$.

81 times (c - a) is a perfect square (because it equals b^2). Since 81 is a perfect square, then (c - a) must also be a perfect square.

01