Calculus Problem

[MHENK]

Find a tough one here

Tough Question

[Guv]

Problem is to cut a wire into pieces and make a parallelopiped with a square base. What shape gives maximum volume? What shape gives maximum surface area?

Cutting the wire into 12 equal pieces and making a cube gives the maximum volume and maximum area.

My intuition suggested that making a cube would maximize the volume, but I was not certain without doing the donkey work to prove it. It was not at all obvious to me that the same shape would maximize both volume and area.

The proof goes something like the following (Sorry if I made a typo in copying from my MathCad Software).

Let a variable X be the decimal percentage amount of wire used for the base and top squares. X must be between zero (none) and one (all of the wire).

Then BasePerimeter = Length*X/2

BaseSide = Length*X/8

HeightWire = Length*(1 - X)

Height = Length*(1 - X) / 4

Volume = Height * BaseSide^2

Area = 2*BaseSide^2 + 4 * BaseSide * Height

Substituting in the above gives.

Volume = X^2 *(1 - X) * Length^3 / 256

Area = X^2 * Length^2 / 32 + X * (1 - X) * Length^2 / 8

Find value of X for which the derivatives are zero.

Derivative( Volume ) = Derivative[ (X^2 - X^3) * Length^3 / 256

Derivative(Volume ) = (2 * X - 3 * X^2 ) * Length^3 / 256

The above will be zero when X * (2 - 3 * X ) = 0

So X = 0, which gives a degenerate shape and a volume of zero.

X = 2 / 3, which gives the value for maximum volume.

From above, BaseSide = Length * X / 8

BaseSide = Length * ( 2/3) * 8

BaseSide = Length(2/24)

BaseSide = Length / 12

Similarly, for the area, find a value of X which makes derivative zero.
Once again, X = 2 / 3 makes derivative of area zero.

It is not really obvious that the cube is correct for both the area and the volume minimization. At least it was not obvious to me until I used my MathCad 7 software to fool with the problem.

[Shaunak]

Because the base is a square one.

Therefore,

8x +4y = 100
y = 25 -2x. .,.... 1

Vol = (x^2)y

substt. the value of y in the vol. from eq 1. and maximise for x.
x = 25/3 and y = 25/3

Similarly for surface area.

[riis]

Shaunak's right.
I'll work it out some more.

x = size of square bottom
y = height

8x + 4y = 100
y = 25 - 2x

Volume: x ^ 2 * y = x ^ 2 ( 25 - 2x) = 25x ^ 2 - 2x ^ 3
Derivate of this should be 0 for extreme value.
Derivate: 50x - 6x ^ 2 = 0 => 50 - 6x = 0 => x = 50 / 6 = 8 1/3

Surface: 2x ^ 2 + 4xy = 2x ^ 2 + 100x - 8x ^ 2 = -6x ^ 2 + 100x
Derivate: -12x + 100 = 0 => x = 100 / 12 = 8 1/3

This (coincidally?) will result in a cube, which seems most logical.

[kedaman]

Well, only mine doesn't assume the base have to be square, but will conclude that there is such symetry.