Largest possible area of the triangle ABC such that AB≤2, BC≤3, AC≤4. Find the largest possible area of the triangle $ABC$ such that $AB$ ≤ 2, $BC$ ≤ 3, and
AC ≤ 4.
Note: The solution should not contain any calculus method
Can someone tell me how to maximize the area of a triangle?
Do we have to make the sides as close as possible to each other to get the maximum area?

Since $AB$ and $BC$ have lower upper bounds, they should be set to maximum. Then if $AC$ is also set to maximum, angle $B$ becomes blunt ($2^2+3^2<4^2$), which is not good. So we set angle $B$ to $90^\circ$ and the maximum area should be $\frac{1}{2}\times 2\times 3=3$. Once the maximum is found (guessed), it's not hard to prove it:
$$S=\frac{1}{2}|AB||BC|\sin B\leq\frac{1}{2}|AB||BC|\leq\frac{1}{2}\times 2\times 3=3.$$
No calculus. Just a simple trigonometry.
