An oil can is to be made in the form of a right circular cylinder to have a volume of 16 pie inches cubed. Find the dimensions of the can that requires the least amount of material
v = pi r^2 h
a = 2 pi r^2 + 2 pi r h
h = 16pi/pi r^2 = 16/r^2
a = 2 pi r^2 + 2 pi r 16/r^2
= 2pi (r^2 + 16/r)
min area when 2r - 16/r^2 = 0
2r^3 - 16 = 0
r = 2
So, the can has a radius of 2 and a height of 4
To find the dimensions of the can that requires the least amount of material, we need to minimize the surface area of the can, as that is directly related to the amount of material needed.
Let's assume the radius of the cylinder is r inches, and the height of the cylinder is h inches.
The volume of a right circular cylinder is given by the formula:
V = πr^2h
Given that the volume of the can is 16π cubic inches, we can write the equation as:
16π = πr^2h
Now, we need to express the surface area of the can in terms of r and h. The surface area of a right circular cylinder is given by the formula:
A = 2πrh + 2πr^2
We want to minimize the surface area, which means we need to find the minimum value of A.
Since we have an equation (16π = πr^2h) that expresses h in terms of r, we can substitute this into the formula for the surface area:
A = 2πrh + 2πr^2
A = 2πr(16π/r^2) + 2πr^2
A = 32π^2/r + 2πr^2
To find the value of r that minimizes the surface area, we need to take the derivative of A with respect to r and set it equal to zero.
dA/dr = -32π^2/r^2 + 4πr
Setting this derivative equal to zero:
-32π^2/r^2 + 4πr = 0
-32π^2 + 4πr^3 = 0
4πr^3 = 32π^2
r^3 = 8π
r = (8π)^(1/3)
Now that we have the value of r, we can substitute it back into the equation for the volume to find the value of h:
16π = πr^2h
16π = π((8π)^(1/3))^2h
16π = π(8π)^(2/3)h
16 = 8^(2/3)h
h = 16 / 8^(2/3)
h = 4 / (2^(2/3))
Therefore, the dimensions of the can that require the least amount of material are:
Radius (r) ≈ (8π)^(1/3) inches
Height (h) ≈ 4 / (2^(2/3)) inches
To find the dimensions of the can that requires the least amount of material, we need to determine the dimensions that minimize the surface area of the can. The volume of a right circular cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius of the base, and h is the height of the cylinder.
In this case, we are given that the volume is 16π cubic inches. So, we can write:
16π = πr^2h
To minimize the surface area, we need to express the surface area in terms of a single variable. The surface area of a right circular cylinder is given by the formula A = 2πrh + 2πr^2, where A is the surface area.
We can rewrite this formula as A = 2πrh + πr^2h/r. Simplifying, we get:
A = 2πrh + πrh = 3πrh
Now, we need to express the height h in terms of the radius r in order to eliminate one variable. From the formula for the volume, we have:
16π = πr^2h
Dividing both sides by πr^2, we get:
h = 16/r^2
Substituting this expression for h into the equation for the surface area, we have:
A = 3πr(16/r^2) = 48/r
To minimize the surface area, we need to find the value of r that minimizes A. To do this, we can take the derivative of A with respect to r, set it equal to zero, and solve for r:
dA/dr = -48/r^2 = 0
Solving, we find that r = √48 = 4√3
Now that we have the value of r, we can substitute it back into the equation h = 16/r^2 to find the corresponding value of h:
h = 16/(4√3)^2 = 16/(48) = 1/3
Therefore, the dimensions of the can that requires the least amount of material are a radius of 4√3 inches and a height of 1/3 inches.