A metal cylindrical container with an open top is to hold 1 cubic

A metal cylindrical container with an open top is to hold 1 cubic foot. If there is no waste in construction, find the dimensions which will require the least amount of material.

A cylindrical metal container with open top is to hold 10 cubic feet of material. If there is

waste in construction, find the dimensions which require the least amount of material.

ahh yes

V = πr^2 h

h = 1/(πr^2)

Surface Area (SA)
= bottom + collar of cylinder
= πr^2 + 2πrh
=πr^2 + 2πr(1/(πr^2)
= πr^2 + 2/r
dSA/dr = 2πr - 2/r^2 = 0 for a min of SA
2πr = 2/r^2
r^3 = 1/π
r = 1/π^(1/3) = appr .693 ft
then h = 1/(π(.693^2)) = .683

Well, I've always been a fan of thinking outside the box, or in this case, outside the cylinder! How about we save some material and make a cylindrical container with a closed top instead? That way, we don't need to worry about constructing an open top and using extra material. Voila! Problem solved. Now go and enjoy your space-saving cylindrical container!

To find the dimensions that will require the least amount of material, we need to minimize the surface area of the cylindrical container while maintaining a volume of 1 cubic foot.

Let's assume the radius of the cylinder is r and the height is h.

The volume of a cylinder is given by the formula V = πr^2h.

We are given that the volume V is 1 cubic foot, so we have πr^2h = 1.

To find the surface area of the cylinder, we need to consider the lateral surface area (the curved surface) and the area of the base.

The lateral surface area of a cylinder is given by the formula A_lateral = 2πrh.

The area of the base is given by the formula A_base = πr^2.

The total surface area of the cylinder is given by A_total = A_lateral + A_base = 2πrh + πr^2.

Now, we need to minimize the surface area A_total, which means finding the dimensions that minimize A_total.

Since we have the constraint πr^2h = 1, we can solve for h in terms of r:

h = 1 / (πr^2).

Substituting this expression for h in the equation for A_total, we have:

A_total = 2πr(1 / (πr^2)) + πr^2 = 2 / r + πr^2.

To minimize A_total, we can take the derivative of A_total with respect to r and set it equal to zero:

dA_total/dr = -2/r^2 + 2πr = 0.

Simplifying this equation, we get:

2πr - 2/r^2 = 0.

Multiplying through by r^2, we have:

2πr^3 - 2 = 0.

Now, we can solve for r:

2πr^3 = 2.

Dividing both sides by 2π, we get:

r^3 = 1 / π.

Taking the cube root of both sides, we have:

r = (1 / π)^(1/3).

Substituting this value of r into the equation for h, we get:

h = 1 / (πr^2) = 1 / (π(1/π)^(2/3)) = (1/π)^(1/3).

Therefore, the dimensions that will require the least amount of material are:

Radius: r = (1 / π)^(1/3).
Height: h = (1/π)^(1/3).

V = π r ² h = 1ft³

h = 1/π r ²

SA = π r ² + 2 π r h
= π r ² + 2 π r * 1/π r ²
= π r ² + 2 / r

SA ' = 2π r - 2 / r² = 0

2π r = 2 / r²
r³ = 1/π

r = 1/ ³√π ft

h = 1/(π r ²)
= 1/(π (1/ ³√π) ²)
= ³√π ² / π
= 1/ ³√π ft