Of all the rectangles....?

Question

having a perimeter of 25 m, find the dimensions of the one with the largest area? What is the largest possible area?
thanks!!!

Answer 1

rectangles have side lengths x and y and perimeter 2x+2y = 25m

area = x.y

(actually the largest such rectangle is a square ... 4x=25, x= 25/4 with area 25^2/16 = 625/16 = 39.0625 sq m)

Lemme work on the proof

2x + 2y = 25
y = 25/2 - x
Area is x(25/2 - x) = 25x/2 - x^2

This equation Area, f(x) = 25x/2 - x^2 describes an inverted parabola (concave down, looks like a frown), the vertex of which is the maximum area (which everyone else has found using calculus - I shall try to keep this simple).

The roots of this parabola can be found readily:
f(x) = x(25/2 - x)

ie = 0 when x = 0 or x = 25/2
Vertical axis of symmetry occurs where x = 25/4

Vertex (ie. maximal height) occurs at x = (25/4), f(x) = (25/4)^2

P (25/4, 625/16)

and the area (maximum height of the function) = 625/16
=39.0625 sq m

Answer 2

Let a and b be the side lengths

The Area = ab

perimeter = 2(a + b) = 25 ==> a = 25/2 - b

Substitute in the area equation to obtain

Area = (25/2 - b)b = 12.5b - b^2

Deriving with respect to b dA/db=12.5 -2b = 0 for a max.

==> b = 12.5/2 = 6.25 m

Like wise from the perimeter equation a = 6.25 m

Thus the area required is 6.25 x 6.25 = 39.06 sq m

Answer 3

if one side is 25/4 and the other side is (3*25)/4 then the area will be maximum. You have to apply differential calculus to solve this.
Basically the function is a*b where a and b are the side lenghts. And the condition you've given is 2(a+b)=25 so substituting for b in the function, differentiate and find a such that the first differential is zero. Also, make sure the second differential is less than zero for it to be maximum at that point.

Answer 4

largest area when each side is 25/4 m that is a square

to prove let sides be a and b

2a+2b = 25

a+b = 12.5

now area = ab

we have 4ab = (a+b)^2 - (a-b)^2

= 12.5^2 -(a-b)^2
maximum when (a-b)^2 is minumum2
maximum when a = b = 6.25

area = 625/16

Answer 5

EASY!
let the dimensions of rectangle are x and y .

area = xy,and 2(x+y) = 25 => x+y=25/2, x= (25-2y)/2

area= (y)(25-2y)/2 (area= yx)

f(y) = y(25-2y)/2 to be max, ( 1st derivative of f(y)=0) for f(y) to be max., y(25-2y) shud be max.

by equating first derivative ( 25-4y ) to zero, we get y=x= 25/4 m =6.25 m(both length and breadth)