Originally Posted by

**Jhevon** .. for the first question, which is kind of wierd:

y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

first we want to know if there are any values we can't pick. remember that square roots are not defined if what is under the square root is negative. since everything under our square roots are squares we have no problem, since any real number squared is greater than or equal to zero always.

so now what would make it a minimum?

consider the first square root:

[(a^2)+(b-x)^2]^(1/2)

a^2 is positive and constant, so we can't do anything about that, but, we can make this value as small as possible if we add 0 to a^2. since adding anything more, we get a bigger value. so we see what's added to a^2 is (b - x)^2. we want this to be zero. in order for (b - x)^2 = 0, x must be equal to b, that way we have (b - b)^2 = (0)^2 = 0.

so for our first root, x = b yields the minimum. for the second root, the same principle applies. we want to add 0 to c^2 in order for it to be a minimum. so we must have x = 0.

so for y to be minimum, we need x = b or x = 0. how do we know which to choose? that depends on the size of a, b, and c