find the economical proportion between the radius and height of the cylindrical can to give the least dimensions of a metal that encloses vloume of 10cu.in
v = πr^2 h = 10
so, h = 10/(Ï€r^2)
The surface area is
a = 2Ï€r(r+h) = 2Ï€r(r + 10/(Ï€r^2))
= 2Ï€r^2 + 20/r
da/dr = 4Ï€ - 20/r^2 = (4Ï€r^2-20)/r^2
da/dr=0 when 4πr^2=20, or r=√(5/π)
so, h = 10/(Ï€*5/Ï€) = 2
Hmmm. I see that my characters for pi and sqrt have been mangled. They appeared correctly as I typed them, but on redisplay are junk.
r = sqrt(5/pi), h=2
Can you re write your solution Steve?
To find the economical proportion between the radius and height of a cylindrical can, we need to consider both the surface area and the volume of the can.
Let's start by finding an expression for the volume of a cylindrical can. The volume (V) of a cylinder is given by the formula:
V = πr^2h,
where r is the radius of the circular base, and h is the height of the cylinder.
Now, let's find an expression for the surface area of the cylindrical can. The surface area (A) of a cylinder is given by the formula:
A = 2πrh + πr^2,
where 2πrh represents the two circular bases, and πr^2 represents the curved surface.
We want to find the economical proportion, so we need to minimize both the surface area and the dimensions of the metal used to enclose a volume of 10 cubic inches.
To find the minimum dimensions, we need to minimize the surface area, which will minimize the amount of metal used. Therefore, we'll differentiate the surface area expression with respect to either the radius or the height and solve for the two variables.
Differentiating with respect to the radius (r):
dA/dr = 2πh + 2πr
Setting dA/dr equal to zero and solving for r:
2πh + 2πr = 0
r = -h
Since we cannot have negative values for the radius, this solution is not valid.
Differentiating with respect to the height (h):
dA/dh = 2πr
Setting dA/dh equal to zero and solving for r:
2πr = 0
r = 0
Again, this solution is not valid as we cannot have a radius of zero.
Since we cannot find a minimum using differentiation, we need to examine the relationship between volume and area. By substituting the volume equation into the surface area equation, we get:
A = 2πrh + V/r
To find the minimum surface area, we set dA/dh equal to zero:
dA/dh = 2πr - V/r^2 = 0
Rearranging the equation, we get:
2πr = V/r^2
Simplifying further:
r^3 = V/(2π)
Taking the cube root of both sides:
r = (V/(2π))^(1/3)
Now, with the value of r, we can find the corresponding value of h by substituting into the volume equation:
V = πr^2h
10 = π[(V/(2π))^(1/3)]^2h
10 = (V/(2π))^(2/3)h
Simplifying further:
h = (10(2π)^(-2/3))/[(V)^(2/3)]
Now, we can substitute the given volume of 10 cubic inches into the equation:
h = (10(2π)^(-2/3))/(10^(2/3))
h = (2π)^(-2/3)
Therefore, the economical proportion between the radius and height of the cylindrical can that gives the least dimensions for enclosing a volume of 10 cubic inches is:
r : h = 1 : (2π)^(-2/3)