Optimization help needed ASAP!?

Question

Sorry to ask such a question, but our study group is at a loss as how to continue and our homework is due tomorrow. So here goes:

In order to receive credit we MUST use calculus techniques:

We have a piece of wire that is 100cm long and we're going to cut it into two pieces. One piece will be bent into a square and the other will be bent into a circle. Determine where the wire should be cut so that the enclosed areas will be at maximum. Note that it is possible to have the whole piece of wire go either to the square or to the circle

Answer 1

Ltotal = 100cm
Lcircle = x cm
Lsquare = (100 - x) cm

now, we want to optimize total area, there are 3 possibilities for a maximum, either end point (i.e., x = 0, or 100), or a local maximum in between.

so first define the area functions and calculate the end points:

Acircle = PI*r^2, Asquare = a^2, where a is the side of the square.

Now we have to define r, and a in terms of x.

The length of the wire for the circle is equivalent to the circumference (C), where C = PI*2*r, thus 2*PI*r = x as we've defined it.

r = x/(2*PI)

for the square, since all 4 sides are equal length, and the total length of the piece cut = Lsquare = 100 - x, the length of the side a = (100-x)/4.

now let's write a function that is the sum of both areas in terms of x, called Atotal(x)

Atotal(x) = PI*(x/(2*PI))^2 + ((100-x)/4)^2

which simplifies to

Atotal(x) = (x^2/(4*PI)) + (100-x)^2/16

now consider the endpoints in this function.

x = 0, Atotal = 625 cm^2
x = 100, Atotal = 795.77 cm^2

Now we check to see if there is a local extrema between them by taking the derivative dAtotal/dx:

dAtotal/dx = x/(2*PI) + (2*(100-x)*(-1))/16

which simplifies to:

dAtotal/dx = x/(2*PI) + (x-100)/8 set this function equal to zero, and solve for x, which is fairly straightforward to get:

x = 43.99

if we plug x = 43.99 into the expression for total area, we find
Atotal(43.99) = 350.062 cm^2, thus, the local extreme must actually a minimum of the summed area. This can be confirmed by a second derivatve test, or visually, by plotting the the total area function.

Thus, the area is maximized by making the entire wire into a circle.

bronco is wrong, he forgot to square the 100-x term when he plugged it back into the area equation.

Answer 2

Let x = the circumference of the circle.
Then (100 - x) = the perimeter of the square
A = π(x/2π)^2 + (25 - x/4)^2
A = (x^2/4π + 625 - (25/2)x - (1/16)x^2
dA/dx = x/2π - 25/2 - x/8 = 0 for min or max
x/π - x/4 = 25
x(4 - π) = 100π
x = 100π/(4 - π)
A'' = 1/2π - 1/8 ≈ 0.0342 > 0, so this point must be a point of minimum area. Since there is only one critical point, the maximum must be an extreme, with all the wire used in either the circle or the square. Intuitively, the circle is the correct choice, but running out the numbers,
square: 25^2 = 625
circle: 10,000/4π = 2,500/π ≈ 795.77 cm^2

Answer 3

To understand what you need to do, figure the area when the wire is used completely for the circle, and then figure the area when the wire is used completely for the square. Whichever is larger is the optimum area you will get; any combination will be between the larger and the smaller of the two.

Answer 4

Ok we want to optimize area,

A = pi*r^2 + s^2

And also, 2*pi*r+ 4s = 100

Now, we let x be the length of the wire to be cut for the square

r = (100-x)/(2*pi)

s = x/4

So now back substituting into the original equation

A = (100-x)/(2) + x^2/16

A = 50 - 0.5x + x^2/16

dA/dx = -0.5 + x/8

Also for optimization, dA/dx = 0

So 0.5 = x/8

And x = 4

So 4 cm should be cut off for the square and that would leave 96 cm for the circle.

I think thats right.