Find the minimum amount of tin sheet that can be made into a closed cylinder havin g a volume of 108 cu. Inches in square inches.
wrong
good catch, Reiny.
To find the minimum amount of tin sheet required to make a closed cylinder with a given volume, we need to consider the surface area of the cylinder.
The volume (V) of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height of the cylinder.
In this case, the volume of the cylinder is given as 108 cubic inches.
To minimize the amount of tin sheet used, we need to minimize the surface area of the cylinder. The surface area of a closed cylinder is given by the formula A = 2πr^2 + 2πrh.
To find the minimum surface area, we can use calculus by taking the derivative of the surface area formula with respect to either r or h, setting it equal to zero, and solving for the minimum value. However, in this case, we can eliminate one variable by expressing the height (h) in terms of the radius (r) using the volume formula.
To do this, rearrange the volume formula to solve for h:
V = πr^2h
h = V / (πr^2)
Now substitute this value for h in the surface area formula:
A = 2πr^2 + 2πr(V / (πr^2))
A = 2πr^2 + 2V / r
To find the minimum surface area, take the derivative of the surface area formula with respect to r:
dA/dr = 4πr - 2V / r^2
Set the derivative equal to zero and solve for r to find the minimum surface area:
4πr - 2V / r^2 = 0
4πr = 2V / r^2
2πr^3 = V
r^3 = V / (2π)
Now substitute the volume value (V = 108) into the equation:
r^3 = 108 / (2π)
r^3 ≈ 17.185
Taking the cube root of both sides gives the approximate value of r:
r ≈ 2.693
Now substitute this value of r back into the height formula to find the height (h):
h = V / (πr^2)
h = 108 / (π(2.693)^2)
h ≈ 4.019
Therefore, the minimum amount of tin sheet required to make a closed cylinder with a volume of 108 cubic inches is approximately 2.693 * 2π * 4.019 square inches.
h = 108/r^2
a = 2pir^2 + 2pi r h
= 2pi r^2 + 2pi r (108/r^2)
= 2pi r^2 + 216pi/r
da/dr = 4pi r - 216pi/r^2
= 4pir^2(r^3-54)
da/dr = 0 when r = 3∛2
a(3∛2) = 2pi(9∛4) + 216pi/3∛2
= (2pi(9∛4)(∛4) + 216pi)/6
= 6pi∛4 + 36pi
= 6pi(∛4+6)
oops in the first line, Steve forgot the π
h = 108/(πr^2)
then
a = 2πr^2 + 216/r
da/dr = 4πr - 216/r^2
= 0 for a min of a
πr = 54/r^2
r^3 = 54/π
r = ∛(54/π) = appr 2.58
then h= 108/(πr^2) = 5.16
notice diameter = 5.16 and height = 5.16