Question

Find the side of the square with vertical and horizontal sides inscribed in the region representing the solution of the system left brace Start 2 By 1 Matrix 1st Row 1st Column y less than minus one half x squared plus 2 2nd Row 1st Column y greater than one half x squared minus 2 EndMatrix

Answer 1

Required side of the squire is
`2.4721`, correct upto four desimal places.

Explanation:

Given two inequalities are,

`y<-(1)/(2)x^2+2\hfill (1)`

`y>(1)/(2)x^2-2\hfill (2)`

which are reflexions of each other across the
`x-`axis. And the square is inscribed within it. Since by symmetry, vertices of the square lies on the lines,

`y=x\hfill (3)`

and
`y=-x\hfill (4)`. We only have to find length of any one side of the squire.

To find
`P`, substitute
`(3)` in
`(1)` we get,

`x^2+2x-4=0`

`x=(-2\pm√(4+16))/(2)=-1\pm √(5)`

Since
`x>0, x=-1+√(5)` and from
`(3)`,
`y=-1+√(5)`, and thus,

`P=(-1+√(5), -1+√(5))`

Similarly by substitute
`(4)` and
`(2)` we will get,

`R=(-1+√(5), -1+√(5))`

And thus,
`\bar{PR}=s=|2√(5)-2|=2.4721`, correct upto four desimal places.