A rectangle has a perimeter of 100cm. What length and width should it have so that its area is a maximum?

Answer 1

p=100cm
p=2(l+b)
for maximum area l*b
100=2(l+b)
or, l+b=50
b=50-l

soA= l*b=l*(50-l)
finding derivative of A
derivative of A= 50-2l
for maximum or minimum value
50-2l=0
l=25
b=25
i.e. total maximum area=625 cm square

Answer 2

Yeah that's a stupid area of maths that i bump into thoroughly retarded because a rectangle may have equivalent aspects like a sq. yet be a rectangle, yet a sq. cant be a sq. except each and each and every of the perimeters are equall.

Answer 3

A square is considered a member of the set that is four-sided polygons and maximizes the area. That set is considered the "recht-angle" ( right-angle in German ) set. ( each side intersects at right angles. A little trivia for your answer.)

Answer 4

lets say L=25+x and B= 25-x.

Area would be: L*B = (25+x)*(25-x)
= 625 -x^2

So it will be maximum when X=0. Then, L and B are both 25 (i.e.) a square!

Answer 5

A square

Proof:
2x + 2y = 100
x + y = 50
y = 50 - x

area = A = x*y = 50x - x^2
dA/dx = 50 - 2x = 0 => x = 25

Answer 6

A square maximizes area, so each side would be 25cm.