Question

A wire four feet long is cut in two pieces. One piece forms a circle of radius r, the other forms a square of side x. Choose r to minimize the sum of their areas. Then choose r to maximize.

How do I do this problem and where do I even start? Very confused! :(

Please explain from step by step? :/ Thanks!! :D

Answer 1

no one else is typing, so i'm gonna have a go, hope that's ok?

Answer 2

yeah sure :)

Answer 3

i honestly have no clue where to start haha

Answer 4

k, circumference of a circle = 2 pi r
area of a circle = pi r^2

circumference of a square = 4s (s is side length)
area of a square = s^2

now we know from question wording that:
Circumference circle + circumference square = 4 ft

we want:
max area
and min area
... with me so far @iheartfood ?

Answer 5

yes:)

Answer 6

sweet
so
Circumference circle + circumference square = 4 ft
2 pi r + 4s = 48 inches

now we want our answer in terms of r, so lets rearrange the above to eliminate s
4s = 48 - (2 pi r)
s = 12 - 0.5 pi r

still cool ? @iheartfood

Answer 7

how did you find the 48 inches part?

Answer 8

1 foot = 12 inches
so 4 ft = 48 inches

Answer 9

ohh okay i see now :)

i understand what you wrote above :)

Answer 10

cool cool,
so now lets start playin with the area's

Area circle + area square = lets call it z (z = total area)
pi r^2 + s^2 = z

and we know that:
s = 12 - 0.5 pi r

so
\[\large pi \times r^2 + s^2 = z\]
\[\large pi \times r^2 + (12 - 0.5 \times pi \times r)^2 = z\]
\[\large pi \times r^2 + 144 - 0.25 \times pi^2 \times r^2 = z\]

so to a few significant figures:
\[\large 0.6742r^2+144 = z\]

would you agree with that?

Answer 11

looks a "r" term is missing...

Answer 12

should it be r^4 ? :/

Answer 13

not quite sure hahaa

Answer 14

damn... so it is, sorry andd good catch @ganeshie8 !! ;)
standby, my bad
do over:
\[\large pi \times r^2 + (12 - 0.5 \times pi \times r)^2 = z\]
\[\large pi \times r^2 + 144 + 2.467r^2 - 37.7r = z \]
\[\large 5.6086r^2 - 37.7r + 144 = z\]

does that look a bit more realistic...?

Answer 15

yes:)

so from here, would we be using quadratic formula? not sure haha <--- just a guess on my part lol

Answer 16

so this equation, if u graph it, is still a parabola, and looks kinda like this:

Answer 17

haha yeah :P

Answer 18

quadratic formula would find where the parabola crosses the x axis
in this case, we want to use derivatives to find where our lowest point is, as the above is a graph of z (area) as a function of r (radius)