You travel the distance from {{MATH_180}} to {{MATH_181}} at speed {{MATH_182}}, and then the distance from {{MATH_183}} to {{MATH_184}} at speed {{MATH_185}}. The distance from {{MATH_186}} to {{MATH_187}} is {{MATH_188}}. By the Pythagorean theorem, the distance from {{MATH_189}} to {{MATH_190}} is {{MATH_191}} Hence the total time for the trip is {{MATH_192}} We want to find the minimum value of {{MATH_193}} when {{MATH_194}} is between 0 and {{MATH_195}}. As usual we set {{MATH_196}} and solve for {{MATH_197}}. Write {{MATH_198}} We find that {{MATH_199}} Notice that {{MATH_200}} does not appear in the last expression, but {{MATH_201}} is not irrelevant, since we are interested only in critical values that are in {{MATH_202}}, and {{MATH_203}} is either in this interval or not. If it is, we can use the second derivative to test it: {{MATH_204}} Since this is always positive there is a local minimum at the critical point, and so it is a global minimum as well.