Question

A piece of wire 40cm long is cut into two pieces. One piece is bent into the shape of a square and the other into the shape of a circle. How should the wire be cut so that the total enclosed is (a) maximum? (b)minimum?

Answer 1

let 40 = s + c

A_square = (s/4)^2
A_circle = pi(c/(2pi))^2

to make a function of 1 variable solve 40 = s + c for either s or c and substitute in the proper area equation.

Answer 2

what do u think it is

Answer 3

you can then max and min using calculus (derivative)

Answer 4

So if I isolate the c, which would be 40-s=c, I would sub it into the area_circle?

Answer 5

yep then the area is the sum of the areas (square and circle)

Answer 6

find dA/ds and set equal to 0. hopefully you have 2 zeros, one which will max and one which will min. but you will have to check

Answer 7

did you figure it out?

Answer 8

I figured out the circle one, which is 40cm.

Answer 9

I didn't quite get the square one...

Answer 10

A = A_square + A_circle

you have to maximize and minimize A not the individual area functions

Answer 11

@pgpilot326 sorry, I didn't quite it...

Answer 12

get*

Answer 13

A = (s/4)^2 + pi((40-s)/2pi)^2

find dA/ds and then find the zeros of dA/ds

check each zero to see if it maximizes or minimizes the function
then you can decide how to cut, accordingly

Answer 14

are you good now or do you have more questions on this?

Answer 15

Yes, I think I got it now. Thanks!

Answer 16

right on. good luck!