# Find the length and width of a rectangle that has the given perimeter and a maximum area. Perimeter: 100 meters

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Expert Answers

violy | Certified Educator

Let set x = width of the rectangle. And y = length of the rectangle.

So, we will have an equation:

2x + 2y = 100 (taking note that the perimter is 100 meters).

Divide both sides by 2.

x + y = 50.

For the area, we will have:

A = xy.

Solve for x using x + y = 50.

Subtract both sides by y.

x = 50 - y.

Plug-in the results on A = xy.

A = (50 - y)(y)

Use the Distributive Property.

A = 50y - y^2.

Differentiate both sides.

A' = 50 - 2y.

Set A' equal to zero.

50 - 2y = 0

Subtract both sides by 50.

-2y = -50

Divide both sides by -2.

y = 25

Solve for x.

x = 50 - 25

x = 25.

Therefore, length = 25 meters, and width = 25 meters.

Student Comments

oldnick | Student

2(x + y) = p x,y sides and p perimeter A = xy
y = (p - 2x)/2 so A =x(p-2x)/2
A = xp/2 - x^2
first find stationary points:
A' = p/2 -2x so A'= 0 if x = p/4
A''= -2 < 0 so the point we found is a max point.
The value is: A= p^2/8 - p^2 /16 = p^2/16 =625 mts^2
the sides are x= 25 mt and y = 25 mt, so the maximum value of area is a square.