1. ## Maths Problems

Post deleted by user.

2. Originally Posted by mathsB
Hi, I'm relatively new to this forum but i hope you will be able to help me with my maths problems.

Problem1.) An ant is at one corner of a cube side length one unit. the ant needs to get from his corner to the corner on the opposite side of the cube (at the top of the cube not the bottom) he must stay on the outside of the cube (on the side or edges) as he cannot fly and the inside of the cube is an open space. What is the least distance he needs to travel to get to the other corner??

N.B. a student came up with a solution of 2.26 using Pythagoras' theorem by folding the side of the cube up and working it out in 2D . Your task is to prove this answer is correct using optimization of calculus.

(Any sort of formula would b a great help)
See attachment.

Now the square of the path length is:

$
l^2=(1+x^2) +((1-x)^2+1)
$

Now the minimum path length corresponds to a minimum of the square path length, so we differentiate $l^2$ with respect to $x$ and set this to zero.

$
\frac{d}{dx}l^2=2x-2(1-x)=4x-2 =0
$

so the required $x=1/2$ and the length is:

$
l=\sqrt{5}
$

RonL

3. Post deleted by user.

4. Post deleted by user.

5. Originally Posted by mathsB
Just one more question. How do u get the equation for L^2?? I know you have to work it in with side X but how do you get the different numbers in there?? And also what geometrical formula do you substitute in?? Any help appreciated
Pythagoras' theorem applied to the two straight line segments on the two faces.

RonL

6. Originally Posted by mathsB
Post deleted by user.
Don't delete your questions, others may benefit from the question and answer. Besides, as your question has been quoted you cannot delete