Question

show that all rectangles with perimeter P the square has the greatest area. i dont understand...

Answer 1

is this algebra or calculus?

Answer 2

its algebra, but it says precalculus honors on the worksheet.

im in the quadratics unit

Answer 3

yes, you can solve it using quadratics

Answer 4

lets say the perimeter is 100
and call one side x and the other y

Answer 5

then \[2x+2=y=100\\
A=xy\] if you solve the first equation for \(y\) you get \(y=50-x\) so
\[A(x)=x(50-x)=50x-x^2\]

Answer 6

damn typo first line should be
\[2x+2y=100\]

Answer 7

ok

Answer 8

the quadratic is therefore
\[A(x)=50x-x^2\] and the max is at the vertex
first coordinate of the vertex is always \(-\frac{b}{2a}\) which in this case is \(-\frac{50}{2\times(-1)}=25\)

Answer 9

so the if the perimeter is 100, then the max is when \(x=25\) i.e. each side is 25 and it is a square

Answer 10

to prove it for all perimeters, replace the \(100\) i used by a \(P\) and repeat the process

Answer 11

so it's simply proving that for any given perimeter of a rectangle, the square's area will be greater than the rectangle?