ElectricPavlov

Yeppers !   Math on......

Area =2x  * ( 4-x)    over the domain  x = 0 to x = 4

Area =  8x - 2x^2

you want to find the maximum value of this function (which is a dome shaped parabola) from x = 0 to x = 4

  you can graph it ...or calculate the max value of x    that produces the max area (the vertex)

     - b /2a = -8/(2* -2) = 2         so x value of vertex = 2      plug this in to the area equation to find y = area = 8 units^2

Or if you want to do the derivative and set = 0

   -4x + 8 = 0

    x = 2     this is where the slope of the parabola is 0 ...this   is the x value which gives the greatest area....the vertex of the parabola....

                     you will have to plug this value of x into the area equation to calculate what the max area is when x = 2

                       8x-2x^2   = y  = area = 8 units²

Here is a graph of the area function:

https://www.desmos.com/calculator/mzd3s0ecs2

The 'domain expressions' are the areas ..or values...of   'x'   that satisfy your equation..

   like from x = 3 to infinity     (  3 < x < + inf)

       or   from   x = 1.5 to 3     which would be  ( 1.5 < x < 3)

In the second graph you can see where the individual functions are + or -   ....

   or where they are both -    or   both +

If you look closely wou will see   +  and -   up until approx x = '1' then a little bit of both + until x ~ 1.5  then   + - again up until 3

       then both +   above 3

   this is graphed in the upper picture  you can see where the division of the two is negative   then a small bit of positive...then negative up to x=3   then positivr above x = 3 

Numerically, there will be different answers.

Example:

-12 * 3 = -36

-12/3 = -4

But I think the domain expressions will be the same.....

Agree!   Sorry....didn't see 'isosceles' in the Q !     ~ EP

Same story, one of the two has to be negative and the other has to be positive .....and neither can be zero.

For GREATER than zero both have to be positive  ....or both have to be negative.

DO you need more than that? 

C_L = L / (A * .5 * r * V² )            L will equal the weight of the plane = 2 897 229 N   

                                                   A = 418 m²

                                                    r=air density   ..  we need the temperature (K)  of the air at 1625 m to calculate density

                                                             using STP and standard lapse rate of 2^o C per 1000 ft results in 4.34^o C = 277.48 K

                                                           pressure = 101.8 kPa    = 101.8 x 10^3  N/m²   = 101.8 x 10^{3  2}kg/(s²-m)  = 10377.17 kg/m²

                                                             using these parameters you can calc   r = 1.27 kg/m³

                                                              V = 99.4 m/s

I find C_L = 1.105      (not positive of this answer)

Multiply first equation by 3 to get

12x +  3x = 15      and add to the 2nd equation to get

 19x = 17

x = 17/19     use this value of x in one of the equations to calc  'y'

Re-arrange to

x^2 - 10x + y^2 + 4y  = -c     now complete the squares for x and y

(x-5)^2      + (y+2)^2   = - c  + 25 + 4          The red portion = r²

-c + 25 + 4 = 1²

29 = 1 + c

28 = c