A rectangular piece of sheet metal with an area of 1200 in2 is to be bent into a cylindrical length of stovepipe having a volume of 600 in3 What are the dimensions of the sheet metal?

Answer 1

The dimensions are approximately 191 inches by 6.28 inches.

This produces a cylinder with about a 2 inch interior diameter, 1 inch radius, and a cross section of 3.14 square inches, 191 inches long.

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Area of the rectangle, A = LW =1200 in^2

where A is also the lateral area of the cylinder L.A.

L.A. = 2(pi)rh

Volume of the cylinder, V = (pi)(r^2)(h) = 600 in^3

Now, let's find the height and the radius of the cylinder.

Let denote W with x, and let W be the height, so h = x. So, L = 1200/x. Let L to be the circumference of the base, so C = L. So we need to find the radius.

C = 2(pi)r

1200/x = 2(pi)r

600/x = (pi)r

r = 600/[(pi)x]

Let's substitute what we know into the volume formula:

V = (pi)(r^2)(h)

600 = (pi)[600/((pi)x)]^2](x)

600 = (pi)[(3600/[(pi)^2)(x^2)](x)

600 = (pi)[3600/(pi)^2)(x^2)](x)

1 = 600/[x(pi)]

x(pi) = 600

x = 600/pi So,

W = x = 191 in (approximately). Then,

L = 1200/191

L = 6.28 in

Or you can work like this: Since,

L.A. = 2(pi)rh = 1200

h = 600/[(pi)r] Substitute h into the volume formula and find r;

V = (pi)(r^2)(h)

600 = (pi)(r^2)[600/((pi)r)]

1 = r substitute r into the height formula;

h = 600/[(pi)r]

h = 600/[(pi)(1)]

h = 600/(pi)

h = 191 So,

h = L = 191 in, and

W = 1200/h = 1200/191

W = 6.28 in