Given a right angle triangle with integral lengths for all 3 sides. Let a, b ,c be the sides of the triangle, and 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.
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.
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.
We want three positive integers where: .Given a right angle triangle with integral lengths for all 3 sides.
Let be the sides of the triangle, and and
Find the maximum area of this triangle under these conditions.
Here's the experimental part . . .
From (b) we have: .
Sustituting (c), we have: .
. . and we see that must be a square.
Since , it narrows down the choices.
I found four solutions: .
Solutions (1) and (2) violate condition (a),
. . which leaves us with solutions (3) and (4).
. .
Hope I didn't miss any . . .
.
Earlier, Soroban said:
First, a minor correction: it should beFrom (b) we have: . a+c)(a-c)" alt="b^2 \:=\:a^2 - c^2 \:=\a+c)(a-c)" />
Sustituting (c), we have: .
. . and we see that must be a square.
c + a)(c - a)" alt="b^2 \:=\:{\color{red}c}^2 - {\color{red}a}^2 \:=\c + a)(c - a)" />
and
.
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.
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