I show you quick a pic from the board in the classroom and you know what i mean

#Extremalwerte

The gaol is, to find the perfect hight and length for the biggest area in the triangel of g, X and Y...
To come at this point we need to get the point "P" wich wasn't there...

P(x|y)
S1(4|0)
S2(0|2 1/3)

First we need to get the equation of g
f(g) = mx + b
f(g) = mx + 2 1/3
now we need m m = y2-y1/x2-x1
f(g) = -(7/12)x + 2 1/3
now we need to get "A" our areavalue
we can use A as Y
A(x) = x(-(7/12)x + 2 1/3)
A(x) = -(7/12)x² + 2 1/3x
YAY we got an "²" wich means?
yeah lets draw a parabola!
but firtst we need to get the point "SP" to draw it, we have our equation in the normal form,
but we cant read the point from it so...
square addition!!!
well i will skip this part and we got the point SP(2|2 1/3)
:/ now some of you think "A = Y" but, if you look at our graphic and at the equation at the top left red box, you see:
A = x * y |:x
A/x = y
(2 1/3)/2 = y
7/6 = y

P(2|7/6) and
x = 2
y = 1 1/6 or 7/6