Find the maximum area of a right triangle with hypotenuse 1m. (Make sure to justify why your answer corresponds to an absolute maximum.)

Here's the answer, but maybe not in a form you were looking for :

Let the above circle have its center at O, and let its radius = 1/2 m.

Then, AC is the diameter of the above circle = 1 m.  Now....  let D be a point on the circle such that OD is  perpendicular to AC and locate B at any point on the circle except at D.

Then let triangle ABC be inscribed in a circle such that its base is a diameter of 1.....and since angle ABC inscribes an arc of 180 degrees......it's measure is 1/2 of this arc = 90 degrees  and AC forms the hypotenuse of this right triangle = 1 m

So.....triangle ABC is a right triangle.......and it's clear that, if we move point B to D, the area of this right triangle ADC will be greater than any other right triangle ABC inscribed in the circle, because  even though their bases will be the same, the height of ADC  will always be greater than the height of any other right triangle ABC.

So the base of ADC = 1m  .....  and its height will be the radius of the circle = 1/2m....so the maximum area of such a triangle will be (1/2)bh = (1/2)(1m)(1/2 m)  = 1/4 m^2......

Here's the Calculus approach.....this may have been answered elsewhere

Let the hypotenuse form the base

And since it's a right triangle, we have that the area = L1 *L2 / 2   where L1 and L2 are the legs

But, by the Pythagorean Theorem L2 =  sqrt( 1 - L1^2)

And let L1  = x

So....the area is given by

A(x) =  [x * sqrt( 1 - x^2 ]  / 2      ....so ....taking the derivative, we have

A' (x) =   (1/2) [ sqrt(1 - x^2)  +  x *(1/2) (1 - x^2)^(-1/2) (-2x)] 

A' (x) = (1/2) [  (1 - x^2)^(1/2) - x^2(1 -x^2)^(-1/2)]  set the derivative to 0

(1/2) [  (1 - x^2)^(1/2) - x^2(1 -x^2)^(-1/2)]  = 0      multiply both sides by 2

 [  (1 - x^2)^(1/2) - x^2(1 -x^2)^(-1/2)]  = 0

[ 1 - x^2]^(-1/2)  [ ( 1 - x^2 - x^2] = 0     multiply through by [ 1 - x^2]^(-1/2) 

[1 - 2x^2]  =  0

2x^2   = 1

x^2  = 1/2

x = 1/√2  = L1

And L2  =   sqrt (1 - L1^2)  = sqrt(1 - x^2)  = sqrt(1 - 1/2)  = sqrt(1/2) = 1/√2

So  ....L1  = L2     and the max area   = (1/2)L1* L2  = (1/2)(1/√2)(1/√2)  = 1/4 m^2

We could take the second derivative to prove this is a max., but it's messy.....much better to graph the area function and note where the max occurs.....here it is :

https://www.desmos.com/calculator/hmmyq9o0io

Note that when x = L1 = L2 = 1/√2     the area is maximized at 1/4 m^2

BTW.....the geometric approach which I presented above seems way more straightforward...!!!