Finding the max volume if you are given surface area?

Question

How would i find the maximum volume if i am given a surface area. For the equation of the surface area I got 64 = 2xy +2xz +2yz. Is that right. And using those variables, i got the equation of the volume to be V = xyz. But where do i go from there?

Answer 1

The equations you mention are appropriate for a rectangular solid whose dimensions are x, y, and z.

If you are trying to find the max volume of A RECTANGULAR SOLID with a particular surface area, then your equations are correct. (If it DOESN'T have to be a rectangular solid, the answer would be a sphere.)

The particular rectangular solid that has the max volume for a given surface area is a cube. You can conclude that by symmetry (i.e., x, y, and z all have to be the same, since the problem is symmetrical with respect to these 3 variables). Although that is not a foolproof method, it gives the right answer in this case, and it is a lot easier than doing the calculus.

Once you know that it's a cube, you just solve for the volume of a cube that has a surface area of 64. (Area of one face = 64/6. Length of an edge = sqrt(64/6) = 8/sqrt(6). Volume = (8/sqr t(6))^3 = 512/(6sqrt(6)) = 128 sqrt(6) / 9.)

If it's not necessarily a rectangular solid, then you use the most symmetrical solid of all: a sphere. If the surface area of a sphere is 64 (= 4 pi r^2), then the radius equals sqrt(16/pi). = 4 / sqrt(pi), and the volume equals (4/3) pi r^3 = (4/3) pi (64/(pi sqrt(pi))) = (256/3) sqrt(pi)/pi.

Answer 2

As it turns out, the shape with largest volume/surface ratio is a sphere. But you are describing a solid parallelopiped, x by y by z.

That is why the volume is xyz and of the six sides, opposite sides are equal 2xy + 2xz + 2yz.

The solid parallelopiped with the best volume/surface ratio is a cube.

So if x = y = z then 64 = 6x**2, which is awkward, to say the least. x = sqrt(64/6)

Then volume = {sqrt(64/6)}**3 = 34.837

as a comparison, consider a box x = 1 and y = 1

64 = 2 + 2z + 2z
z = (31) / 2

and volume = 15.5

Answer 3

If V your volume is equal to xyz then you have a rectangle of lenght x by width y by height z.
Now the surface of a rectangle if xy + xy + xz + xz + yz + yz which are the six planes of your rectangle.
Thus you get 2xy + 2xz + 2yz and in your case gives you 64
Simplifying this out you get xy + xz + yz = 32
We would get a maximum volume if x=y=z
So we would get 3x*2 = 32
x*2= 32/3
or x=y=z = square root 32/3
V= xyz= x*3
V=[square root 32/3]*3
V=32/3*3/2
V=34.84

Answer 4

In case you want to prove that a cube has the max volume for a given rectangular solid, use Lagrange multipliers. You want,

gradient V = constant * gradient (constraint)
gradient (xyz) = constant * gradient (2*x*y + 2*x*z + 2*y*z - 64)
(y*z, x*z, x*y) = 2*constant * (y+z, x+z, x+y) = L * (y+z, x+z, x+y)
which gives the 3 equations,
y*z = L * (y+z)
x*z = L * (x+z)
x*y = L * (x+y)

You can see by inspection that x=y=z and L = x/2 solves the equations.

Answer 5

Without going into the proof, maximum volume of a rectangular solid occurs when x=y=z, so,
64 = 6x^2
x = 8/sqrt(6)
V = x^3 = 512/6^(3/2) = 34.837

However, the maximum volume of a given surface area is a sphere:
A = 4πr^2
V = (4πr^3)/3 = (Asqrt(A/4π))/3
V = (64sqrt(16/π))/3
V = 48.144

Answer 6

with a bit of luck, this could be a calculus concern. the quantity of a cylinder is the area of the tip cases the top. (a million) V = pi * r^2 * h And the exterior area is the two ends, plus the area that unrolls right into a rectangle: (2) area = 2 * pi * r^2 + 2*pi*r*h = 80 pi From (2), you are able to clean up for h in terms of r. Then, replace this fee for h into (a million). you presently have an equation for the quantity in terms of r in basic terms. you are able to maximize this quantity by utilising taking the by-product and placing it to 0, then taking the 2d by-product and making optimistic that the curve is concave down at that ingredient.

Answer 7

T = 2(ab + ac + bc)

if you have it correct, then using the same concept

64 = 2(xy + xz + yz)
xy + xz + yz = 32

xy + yz + xz = 32
y(x + z) + xz = 32
y(x + z) = -xz + 32
y = (-xz + 32)/(x + z)

V = x * ((-xz + 32)/(x + z)) * z
V = (-x^2z^2 + 32xz)/(x + z)

thats about all i can figure out.

if x = y, then

xy + yz + xz = 32
x^2 + zx + zx = 32
x^2 + 2xz = 32
x^2 + 2zx - 32 = 0

x = (-2z ± sqrt((2z)^2 - 4(1)(-32)))/(2(1))
x = (-2z ± sqrt(4z^2 + 128))/2
x = (-2z ± sqrt(4(z^2 + 32)))/2
x = (-2z ± 2sqrt(z^2 + 32))/2

x = -z + sqrt(z^2 + 32)

therefore

V = x^2 * (-z + sqrt(z^2 + 32))

so it all depends on somethings.