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

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violy | High School Teacher | (Level 1) Associate Educator

Posted on

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. 

 

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oldnick | (Level 1) Valedictorian

Posted on

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.

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