Question

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

Accepted Answer

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.

Answer

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.

Math

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

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