3 ms·
It's y=(sqrt(x) - 1)^2 Consider the function l(t) which gives the appropriate line function f(x) for the line going through (t,0). The function we are looking
by chnuschper 13y ago
It's y=(sqrt(x) - 1)^2
Consider the function l(t) which gives the appropriate line function f(x) for the line going through (t,0). The function we are looking for can then be defined as c(x) := max(l(t)(x), t in (0;1]). Just solve for t and pop it into l(t), et voila.
- jules 13y agoSorry, I accidentally down-voted you when I tried to up-vote you. This answer is absolutely correct. Here is my very similar derivation: All lines go through (t,0) and (0,1-t). Now take two such lines, one with constant t and another with constant s and find their intersection. We have: y = 1-t - (1-t)/t * x y = 1-s - (1-s)/s * x So: 1-t - (1-t)/t * x = 1-s - (1-s)/s s-t = (s-t)/(st) * x So: x = st The points on the curve will be generated by the intersection of two almost adjacent lines (this is easy to see geometrically), so we take s =~ t. Then we have: x = t^2 y = (1-t)^2 So we have: y = (1 - sqrt(x))^2 Or, my favorite form: sqrt(x) + sqrt(y) = 1