#1

I need some calculus help, UG. I'm really stuck on a couple questions. If someone would help me through one or both of them, that would be great.

1. You are in a vehicle in the desert 12 km due south of the nearest point A on a straight east-west road. You wish to get to point B on the road 10 km due east of A. If the vehicle travels 15 km/h over desert, and 39 km/h on the road, toward what point on the road, x km from A, should you head in order to minimize travel time to B?

2. A box with an open top is to be constructed from a piece of cardboard, 1m by 2m, by cutting out a square piece from each of the four corners and bending up to make the sides. Find the largest volume that such a box can have.

1. You are in a vehicle in the desert 12 km due south of the nearest point A on a straight east-west road. You wish to get to point B on the road 10 km due east of A. If the vehicle travels 15 km/h over desert, and 39 km/h on the road, toward what point on the road, x km from A, should you head in order to minimize travel time to B?

2. A box with an open top is to be constructed from a piece of cardboard, 1m by 2m, by cutting out a square piece from each of the four corners and bending up to make the sides. Find the largest volume that such a box can have.

#2

Ha... This is funny

Alright, right now you have two variables (the distance driven on the road, and the distance driven on the desert). What you need to do is get one variable in terms of the other. ie. get your velocity equation in terms of only one variable.

If we let x=distance driven on the road, then we need to get the distance driven on the desert in terms of x as well. To do this, we need to use a little trig.

If you draw this out, you can see the distance driven on the desert is the hypotenuse of a triangle. one side has length 10km, the second is (12-x)km. You now have the second distance in terms of x.

Now, set up this equation

f = 39x + 15(SQRT( 10^2 +(12-x)^2))

Hopefully you can see why this equation is used (I think I could only explain properly in person).

Anyway, now you have to take the derivative of that equation and set it to zero (to find a local maximum) and solve for x.

When you have x, you will have your answer

Alright, right now you have two variables (the distance driven on the road, and the distance driven on the desert). What you need to do is get one variable in terms of the other. ie. get your velocity equation in terms of only one variable.

If we let x=distance driven on the road, then we need to get the distance driven on the desert in terms of x as well. To do this, we need to use a little trig.

If you draw this out, you can see the distance driven on the desert is the hypotenuse of a triangle. one side has length 10km, the second is (12-x)km. You now have the second distance in terms of x.

Now, set up this equation

f = 39x + 15(SQRT( 10^2 +(12-x)^2))

Hopefully you can see why this equation is used (I think I could only explain properly in person).

Anyway, now you have to take the derivative of that equation and set it to zero (to find a local maximum) and solve for x.

When you have x, you will have your answer