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?

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
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Area of a square = 75^2 = 5625 m^2 = maximum
.
To find the minimum area...
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Assume the rectangle is 1x149.
Perimeter = 2(1+149) = 2(150) = 300 m.
Area = 1*149 = 149 m^2
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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.
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Assume 'x' is 1 cm = .01 m. Then 'y' will be 150-.01 = 149.99 m.
.01*149.99 = 1.4999 m^2
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Assume 'x' is 1 mm = .001 m. Then 'y' will be 150-.001 = 149.999
Area = .001*149.999 = .149999 m^2
etc.
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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.

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