You need to notice that the problem provides the rate the diameter of the spill increases such that:

`(dD)/(dt) = 1/4` miles/hour

You need to write the area of the circle in terms of diameter such that:

`A = pi*D^2/4`

You need to differentiate the area with respect to t such that:

`(dA)/(dt) = 2piD/4*(dD)/(dt) `

You need to evaluate the diameter if `A = 100` square miles such that:

`A = pi*D^2/4 =gt 100 = pi*D^2/4 =gt 400 = pi*D^2`

`200/sqrt pi= D`

You need to substitute `200/sqrt pi` for D and `1/4` for `(dD)/(dt)` in `(dA)/(dt) = 2piD/4*(dD)/(dt)` such that:

`(dA)/(dt) = 2pi 200/(4sqrt pi)*(1/4)`

`(dA)/(dt) = 25 sqrt pi` square miles per hour.

**Hence, evaluating how fast the area of the spill increasing under given conditions yields `(dA)/(dt) = 25 sqrt pi` square miles per hour.**