If the diagonal of the square is `x` , then a side of the square, `a` is,

`a = x cos(pi/4) = x/sqrt(2)`

The perimeter, `y` is given by,

`y = 4a`

`y = 4x/sqrt(2)`

`y = 2sqrt(2)x`

If the area is 18, then,

`a^2 = 18`

`(x/sqrt(2))^2 =18`

`x^2/2...

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If the diagonal of the square is `x` , then a side of the square, `a` is,

`a = x cos(pi/4) = x/sqrt(2)`

The perimeter, `y` is given by,

`y = 4a`

`y = 4x/sqrt(2)`

`y = 2sqrt(2)x`

If the area is 18, then,

`a^2 = 18`

`(x/sqrt(2))^2 =18`

`x^2/2 =18`

`x =6`

Therefore when `x =6, (dx)/(dt) = 3`

We have to find the rate of increase of perimeter, `(dy)/(dt)`

`y = 2sqrt(2)x`

`(dy)/(dt) = 2sqrt(2) (dx)/(dt)`

`(dy)/(dt) = 2sqrt(2) xx 3`

`(dy)/(dt) = 6sqrt(2)`

**Therefore the rate of increase of perimeter is** `6sqrt(2)` ** inches per minute.**