Question

can anyone help?
show that of all the rectangles with a given area, the one with the smallest perimeter is a sqaure

Answer 1

i dont understand

Answer 2

are u allowed to use calculus?

Answer 3

yes

Answer 4

ok....
so let P=2x+2y, where the rectangle has length x, and width y...

Answer 5

so we need to minimize P....

Answer 6

area of the rectangle is given, so A=xy .
this is our secondary equation that we'll need to substitute into the primary equation (perimeter equation)

Answer 7

solving for y, we have y=A/x

Answer 8

substituting into the perimeter equation we have:
P = 2x + 2(A/x)
\(\large P = 2x + 2Ax^{-1} \)

now you differentiate P....

Answer 9

if i continue from where u are i have p'=a

Answer 10

no... you did not differentiate correctly...

Answer 11

P'=\[\pm \sqrt{A}\]

Answer 12

sorry X not P'

Answer 13

\(\large P=2x+2Ax^{-1} \)
\(\large P'=2-2Ax^{-2} \)

Answer 14

now set that derivative equal to zero and solve for x....

Answer 15

alright....i see it now

Answer 16

thanks

Answer 17

\(\large P'=2-2Ax^{-2} \)
\(\large 0=2-2Ax^{-2} \)
\(\large 0=1-Ax^{-2} \)
\(\large Ax^{-2}=1 \)
\(\large x^{-2}=\frac{1}{A} \)
\(\large x^{2}=\frac{A}{1} =A\)
\(\large x=\sqrt{A} \)

since x = squareroot A,
y must be squareroot A also...
in other words, x=y....

Answer 18

since x=y, the rectangle is a square...

Answer 19

thanks

Answer 20

yw... :)