Ask Me Help Desk - Quadratic: Rectangle

Quote:

Originally Posted by jesse smith

A rectangular nuclear-waste holding facility is 100 m long and 70 m wide. A safety zone of a uniform strip of more than 20 m in width must be constructed around the facility. Determine the width of the strip of land to the nearest metre, if the area of the facility and the strip is 90 000 m^2.

The area of the facility is $100 m * 70 m = 7000 m^2$

The total area, as given, is $90000 m^2$ so the area of the strip must be $90000 m^2- 7000m^2 = 83000 m^2$

If the safety zone were 20 meters wide, that would be a rectangle 140 m long and 110 m wide. That's the minimum. I'm not actually sure what part that plays in the problem.

$140 x 110 = 15400 m^2$

The width of the facility + safety zone can be expressed as follows

$(100 + 2W) (70 + 2W) = 83000$ where W is the width of the safety strip.

Solving for W

$7000 + 340 W + 4W^2 = 83000$

and rearranging

$4W^2 + 340W - 76000 = 0$

$W^2 + 85W - 19000 = 0$

$W = -85$ ± $\sqrt{7225 + 76000}$

$W = -85$ ± $288.48$

Since the negative root has no meaning,
W=203.48 m

Unless I made a math mistake along the way.