how would u show that among all rectangles w/ an 8-m perimeter, the one w/ largest area is a square?

Answer 1

Say that the dimensions of the rectangle are x and y. Then

2x+2y=8 and A=xy

Simplify and solve the first for x to get
x+y=4
x=4-y

Substitute this into the second:

A = (4-y)(y)
A = 4y-y^2

The x-coordinate of the parabola is -b/2a, where a is the x^2 coefficient and b is the x coefficient. Here, a=-1 and b=4, so the x-coordinate of the vertex is -(4)/2(-1) = 4/2 = 2.

Obviously, if x=2, then y=2 and A=4. The vertex is the maximum of the parabola, all of the other values are smaller than it. So, given a fixed perimeter of a rectangle, the largest area will always be when the figure is a square

Answer 2

in the case of a squarewith a perimeter of 8 m,each side is
8/4=2 m
Area is 2*2=4 sq,m
if it is some other rectangle the sides would be 2+x and 2-x
Area=(2+x)(2-x)
=4-x^2 [(a+b)(a-b)=a^2-b^2]
Whatever may be the value of x,4-x^2 is always less than 4
Hence the area of a square is always morethan any other rectangle,whenever they have the same perimeter

Answer 3

2(x+y)=8(perimeter ) so x+y= 4

S=xy max S= x*(4-x) = -x^2+4x = -(x-2)^2 +4.

The maximum is for x=2 because -(x-2)^2 is negative for all other x
So x=2 and y= 4-x=2 It's a square