What is the maximum ha (vertical height) of the ABC triangle given its perimeter = 19 m and hb = 7 m (lateral height)?
Not sure what hb means. If hb is the lateral height, that is one of the sloping sides of the triangle, where is point H? Also, is ha the altitude to side a? Very unclear.
So, I'll answer a question of my own devising. Suppose one of the sloping sides has length 7. Then if the base is AB, of length x, and the altitude from C to AB meets AB at a distance y from vertex A, and the other sloping side is of length z, then I suspect maximum height will be when AC=BC = 7 and AB=5.
Let's see what happens. If the altitude is h, then
x+z+7 = 19, so x+z=12
y^2+h^2 = z^2
(x-y)^2 + h^2 = 7^2
h^2 = z^2-y^2 = 49-(x-y)^2
(12-x)^2 - y^2 = 49-(x-y)^2
y = x - 12 + 95/(2x)
dy/dx = 1 - 95/(2x^2)
dy/dx=0 when x = √(95/2) = 6.892
Then y = 1.784
and maximum h = 4.786
and the third side z = 5.108