We have just started studying functions of several variables and their derivatives and our professor suggested the following problem as food for thought. Two squares, both with length l=1 intersect in a rectangle that has an area equal with 1/8 . Find the minimum and maximum distance between the centers of the squares.

Austin Rangel

Austin Rangel

Answered question

2022-10-06

We have just started studying functions of several variables and their derivatives and our professor suggested the following problem as food for thought.
Two squares, both with length l=1 intersect in a rectangle that has an area equal with 1/8 . Find the minimum and maximum distance between the centers of the squares.

Answer & Explanation

omeopata25

omeopata25

Beginner2022-10-07Added 5 answers

You need to introduce some coordinates in there to transform that geometry problem into an algebra one.
So let's fix the first square between ( 0 , 0 ) and ( 1 , 1 ), so with center ( 1 / 2 , 1 / 2 ). Next, you have ( x , y ) the coordinates of one edge of the other square, which defines that square up to 4 cases. We will focus on ( x , y ) being the top right corner.
The center of that second square is then ( x 1 / 2 , y 1 / 2 ). The distance between the 2 centers is then d ( x , y ) = ( x 1 ) 2 + ( y 1 ) 2
So now the problem is to find the extremum of d ( x , y ) (or better d 2 ) subject to x y = 1 / 8.
With Lagrange multipliers, that's finding the extremum of ϕ ( x , y , λ ) = ( x 1 ) 2 + ( y 1 ) 2 λ ( x y 1 / 8 ) such that ϕ x = ϕ y = 0 = 2 ( x 1 ) λ y = 2 ( y 1 ) λ x
that implies x = y = 1 2 2 (easy to verify it's a minimum by looking at the 2nd derivative).
As for the maximum, it's obviously not a critical point, so has to be on the border. For instance x = 1 and y = 1 / 8.

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