I tried solving the question in this post in another way and got a different result,
Let $0\lt a\lt b$
(i) Show that among the triangles with base $a$ and perimeter $a + b$, the maximum area is obtained when the other two sides have equal length $b/2$.
My idea was to treat the side as vectors of same length, so $\vec{a}$ be one side and $ \vec{p}$ be an adjacent side, both vectors originating from the same point, then the maximum area occurs when both of them are perpendicular. (Note: $\vec{a}$ represents the side of side length $a$
Proof: $ \vec{a} \times \vec{p} = |a| |p| \sin \theta$, expression is maximized at $ \theta = \frac{\pi}{2}$, so this suggests the right angle at the common origin of $\vec{a} $ and $\vec{p}$
However, if this the case the perimeter condition can never be satisfied.. What went wrong here?
Best Answer
You have assumed that the length of one side given by $\vec{p}$ is fixed, but that would give no further freedom as the length of the other side will also be fixed due to the perimeter constraint, and so the angle between $\vec a$ and $\vec p$ is fixed. To do the problem, you need to assume exactly one degree of freedom - either $\theta$ or $|\vec p|$.