The dimensions of the sheet metal can be any combination of width (w) and length (l) that satisfy the following equations:
1200 = w × l
1200w = 2400πr - 4πr²w
Let's denote the width of the rectangular sheet metal as 'w' and the length as 'l'.
The area of the rectangular sheet metal is given by:
Area = w × l
We are given that the area is 1200 in², so we can write:
1200 = w × l
We are also given that the volume of the cylindrical stovepipe is 600 in³. The volume of a cylinder is given by:
Volume = πr²h
where r is the radius of the base and h is the height.
Since the sheet metal is to be bent into a cylindrical stovepipe, we have the following relationship:
2πr × h = 1200
We can also express the height (h) in terms of the width (w) and length (l) of the rectangular sheet metal:
h = l - 2w
Substituting this expression for h into the volume equation, we get:
2πr × (l - 2w) = 1200
Now we have two equations with two unknowns (w and l). We can solve for w and l using substitution or elimination. Let's use substitution.
From the area equation, we can express l in terms of w:
l = 1200 / w
Substitute this expression for l in the volume equation:
2πr × (1200 / w - 2w) = 1200
Simplifying and rearranging:
2400πr - 4πrw² = 1200w
Dividing both sides by w:
2400πr / w - 4πr² = 1200
This gives us an equation in terms of w. We can solve for w numerically or use a graphing calculator to find the values of w that satisfy the equation.