Find the dimensions of a rectangle with perimeter 100m. whose area is as large as possible.

Answer 1

x + y = 50
so
y = 50 -x

A = x y = x(50 - x ) = 50 x - x^2
dA/dx = 0 for max or min = 50 -2x
2 x = 50
x = 25
so it is a square 25 on a side.

Answer 2

thanks a lot!

Answer 3

let x be width and y be length ,and as we know perimeter of rectangle=2(length+widith)
so 2(x+y)=100
x+y=50 , x=50-y
area=length*width
area=xy=(50-y)y
area=50y-y^2
dA/dy=0 for min or max
dA/dy=50-2x=0
y=25
x=25

Answer 4

let x be width and y be length ,and as we know perimeter of rectangle=2(length+width)
so 2(x+y)=100
x+y=50 , x=50-y
area=length*width
area=xy=(50-y)y
area=50y-y^2
dA/dy=0 for min or max
dA/dy=50-2y=0
y=25
x=25

Answer 5

Thank you so much