A container open at the top is a right circular cylinder having a surface area of 108 cm^3. What should the radius and altitude be in order to provide the largest possible volume?

So the surface area is made up of the circular base and the "sleeve" of the cylinder.

πr^2 + 2πrh = 108
h = (108 - πr^2)/(2πr) = 59/(πr) - r/2

V = π r^2 h
= πr^2(59/(πr) - r/2)
= 59r - (π/2)r^3
dV/dr = 59 - (3π/2)r^2 = 0 for a max of V

(3π/2)r^2 = 59
solve for r, then get h

To find the dimensions that would provide the largest possible volume for the container, we need to consider the relationship between the surface area and the volume of a right circular cylinder.

Let's call the radius of the cylinder "r" and the altitude (height) of the cylinder "h".

The surface area of the cylinder is given by the formula:

Surface Area = 2πr² + 2πrh

We are given that the surface area is 108 cm², so we can write the equation as:

2πr² + 2πrh = 108

Now, the volume of the cylinder is given by the formula:

Volume = πr²h

Our goal is to find the dimensions that maximize the volume. To do this, we can solve the equation for h in terms of r, substitute it into the volume formula, and then differentiate the volume equation with respect to r to find where the derivative is equal to zero.

First, solve the surface area equation for h:

2πrh = 108 - 2πr²
h = (108 - 2πr²) / (2πr)

Substitute this expression for h into the volume equation:

Volume = πr² ((108 - 2πr²) / (2πr))

Simplifying the expression, we get:

Volume = (54r - πr³)

To find the radius that would provide the maximum volume, we differentiate the volume equation with respect to r and set it equal to zero:

d(Volume)/dr = 54 - 3πr² = 0

Solving for r:

3πr² = 54
r² = 18 / π
r = sqrt(18 / π)

Substituting this value of r back into the surface area equation, we can solve for h:

2πr + 2πrh = 108
2π(sqrt(18 / π)) + 2π(sqrt(18 / π))h = 108
sqrt(18 / π) + sqrt(18 / π) * h = 54 / π
(1 + h) * sqrt(18 / π) = 54 / π
h = (54 / π) - sqrt(18 / π)

Therefore, the radius that would provide the largest possible volume is sqrt(18 / π) and the altitude would be (54 / π) - sqrt(18 / π).

To find the dimensions of the right circular cylinder that would provide the largest possible volume, we need to use calculus.

Let's assume that the radius of the cylinder is 'r' and the altitude (height) is 'h'.

The surface area of a right circular cylinder is given by the formula:
A = 2πr^2 + 2πrh

The problem states that the surface area is 108 cm^3, so we can write the equation as:
108 = 2πr^2 + 2πrh

To find the maximum volume, we need to maximize the function V = πr^2h. However, we have two variables (r and h), so we need an additional equation to solve for one of the variables in terms of the other.

Since the cylinder is open at the top, the altitude cannot be greater than the radius, otherwise, it would not be a cylinder. So we can assume h ≤ r.

Now, we need to express h in terms of r. We can rearrange the surface area equation for h as follows:
h = (108 - 2πr^2) / (2πr)

Substituting this expression for h into the volume equation, we get:
V(r) = πr^2 * [(108 - 2πr^2) / (2πr)]

Simplifying this equation, we get:
V(r) = (54r - πr^3) / 2

Now, we can differentiate V(r) with respect to r and set the derivative equal to zero to find the critical points. We can then check the second derivative to ensure that the critical point corresponds to a maximum.

Taking the derivative of V(r) with respect to r, we get:
V'(r) = 54/2 - 3πr^2/2

Setting V'(r) equal to zero, we solve for r:
54/2 - 3πr^2/2 = 0
27 - 3πr^2 = 0
3πr^2 = 27
r^2 = 9/π
r = √(9/π)
r = 3/√π

Now, we need to check the second derivative, which is:
V''(r) = -6πr

Substituting the value of r found above, we get:
V''(3/√π) = -6π(3/√π) = -18

Since V''(r) is negative, it confirms that the critical point r = 3/√π corresponds to a maximum volume.

Therefore, the radius 'r' that would result in the largest possible volume is 3/√π, and the altitude 'h' would be less than or equal to this value to maintain the shape of a cylinder.