# SOLUTION: what must be the value of x and y if the area is to be a minimum and the perimeter is to be 300m?

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 Question 515755: what must be the value of x and y if the area is to be a minimum and the perimeter is to be 300m?Answer by oberobic(2304)   (Show Source): You can put this solution on YOUR website!x = length y = width . To maximize the area, use a square. To minimize the area, decrease x and increase y by the same amount. . The perimeter = 300 (a constant) 2(x+y) = 300 x+y = 150 x = 150-y y = 150-x . Area of a square = 75^2 = 5625 m^2 = maximum . To find the minimum area... . Assume the rectangle is 1x149. Perimeter = 2(1+149) = 2(150) = 300 m. Area = 1*149 = 149 m^2 . Assuming 'x' can be made infinitely small, then it can practically disappear. As 'x' gets closer and closer to 0, the 'y' gets closer and closer to 150 m. Recall y = 150-x. . Assume 'x' is 1 cm = .01 m. Then 'y' will be 150-.01 = 149.99 m. .01*149.99 = 1.4999 m^2 . Assume 'x' is 1 mm = .001 m. Then 'y' will be 150-.001 = 149.999 Area = .001*149.999 = .149999 m^2 etc. . As you may have guessed. the limit of the area as 'x' approaches 0 is 0 m^2. That would the case where the fence was doubled back on itself with no area at all. The perimeter would still be 300, though it would look like a fence line 150 m long that was double thickness.