Question

Please explain, i am getting smaller area

Answer 1

so 100 x 130 = 13000

Answer 2

115 x 115 = 13225

Answer 3

im not sure what the secret rule is here but, actually that was a guess.

Answer 4

its correct ...

Answer 5

given the sum of the numbers the product is maximum if the numbers are individully half of the sum...

Answer 6

i formed an inequality but got less...i think if i change the inequality sign i will get more

Answer 7

11 x 9 = 99
10 x 10 = 100

same perimeter again

Answer 8

ya

Answer 9

interesting though

Answer 10

so you are using trial and error?

Answer 11

no

Answer 12

i got the 115, by subtracting 15 from 130 and adding it to 100

Answer 13

can be proved using calculus...

Answer 14

(100+130)/2

Answer 15

ya the average

Answer 16

make a new rectangle that is closer to a square in shape

Answer 17

it cant be...

Answer 18

just took the perimeter of 230

2piR = 230
230/2pi = r

A=piR^2 = 1344.86

Answer 19

hmmm....this is interesting ...so you used the average?

Answer 20

ya thats the best way to do it

Answer 21

turn it into a square

Answer 22

ok...hold that thought. let me give you a similar question, with a little twist though

Answer 23

here it is

Answer 24

60 x 105 = 6300

330 = perimeter

Answer 25

uh..you should just form an equation for perimeter with lets say x and y...then form an inequality for area...that works for both instances

Answer 26

115 x 50 = 5750

Answer 27

in this case i just made it less like a square

Answer 28

shaved a 10 off of 60 added it to 105

Answer 29

hmm.... i see how you did it now

Answer 30

but there must be a formula to it

Answer 31

wait a ^(#)^(#ing minute here

Answer 32

1 x 164

Answer 33

is as small as you can get it with the same perimeter 330

Answer 34

you could probably come up with an equation for it

but its just arithmetic really.

Answer 35

lol ... i wouldnt have guessed that

Answer 36

yeah...i guess it is

Answer 37

it is not trivial to show that if you hold the perimeter constant, that the shape that gives the max area is a square, and the shape that gives min area is a rectangle that is *almost* a thin line. (very very long and very short height)

Answer 38

would it reach a point where one side is so small that the area actually becomes bigger?

Answer 39

if we let x+y= P (where P stands for ½ of the perimeter) then
y= P-x
and area A= x*y= x(P-x)

A= -x^2 + xP

do you know how to "complete the square" ?
Write this as
\[ A= - ( x^2 - xP + \frac{P^2}{4} - \frac{P^2}{4} ) \]
(I factored out a minus sign. I added and subtracted P^2/4 (so nothing changes)
But we now have a perfect square
\[ A= -\left ( \left (x - \frac{P}{2} \right)^2- \frac{P^2}{4} \right) \]
distribute the minus sign:
\[ A= \frac{P^2}{4} - \left (x - \frac{P}{2} \right)^2\]

now we notice that the area is p^2/4 minus a number (the messy stuff squared)
that squared number is *never* negative. It is either 0 or bigger.
if it is 0, then the area will be the biggest possible.

In other words, the biggest area is when
\[ \left (x - \frac{P}{2} \right)^2 = 0 \\ x-\frac{P}{2} =0 \\ x= \frac{P}{2} \]

and using y= P-x, when y= \(\frac{P}{2}\)
i.e. when x=y= \(\frac{P}{2}\) , a square

Answer 40

@cece1902

Answer 41

sorry
what is this about

Answer 42

phi = god