More optimization problems

Hello everyone this is the scribe post for May 6, sorry it is late but I had to write the AP calc exam and I was not able to post it earlier. Well in Tuesday's class we only did one question, but there were two others for homework. The first question was about a man who was driving his car through the desert, and wanted to know at what distance he should drive through the desert and the paved road to get to the city B 5km south from city A. He can drive 15mph in the desert and 35mph on the paved road, he is 5 km east of city A. So the maximum distance the man could travel is 5km in the desert and 5km of pavement to get to city B (The diagram is on the slides). So that would be unrealistic, so in order to get the minimum amount of time the car must cut time through the desert and pavement alike. So the pavement equation will become 5km subtract x, x the amount of distance that is going to be cut by driving through the desert. Then the amount of distance driving through the desert is found by a^2 + b^2 = c^2. Where a and b is 5-x. So c is (25+x^2)^.5 (square root is ^.5).

As you know Rate, Distance, and Time is all related. Time is Distance over rate. So the total time is (5-x)/35 + ((25+x^2)^.5)/15 =t(x). Solving for the derivative (solved on the slides) , t'(x) =
7x-3((25+x^2)^0.5)/105((25-x^2)^0.5). Now solving for the zero's you get 40x^2-225. So the roots are -15/2((10)^0.5) and 15/2((10)^0.5). We reject the negative value because its not possible in the context of the question. So the man should go and head for the point x = 15/2((10)^0.5), as that point is the minimum value of the First Derivative Function.

Remember to do the other two questions and the homework given in class, the next scribe is Ethan.