Find the economical proportion between the radius and height of the cylindrical can to give the
least dimensions of a metal that encloses a volume of 10 cu.
Let the radius of the can be r and its height be h
V = π r^2 h
10 = πr^2h or h = 10/(πr^2)
For your least dimension, the surface area (SA) has to be a minimum
SA = 2πr^2 + 2πrh
= 2πr^2 + 2πr(10/πr^2) = 2πr^2 + 20/r
D(SA)/dr = 4πr - 20/r^2 = 0 for a minimum of SA
4πr = 20/r^2
r^3 = 5/π = 1.591549
h = approx 1.1675
sub this back into h
h = approx 2.335
which is twice the value of r
So the height should be twice the radius
or
for a minimum SA , h : r = 2 : 1
(the height should equal the diameter)
To find the economical proportion between the radius and height of the cylindrical can, we need to minimize the surface area of the can while keeping its volume at 10 cubic units.
Let's assume the radius of the cylindrical can is "r" and the height is "h".
The volume of a cylindrical can is given by the formula V = π * r^2 * h.
We are given that the volume should be 10 cubic units, so:
10 = π * r^2 * h
Now, we need to minimize the surface area of the can, which is the sum of the lateral surface area and the top and bottom areas.
The lateral surface area of a cylindrical can is given by the formula A_lateral = 2πrh.
The top and bottom areas are given by A_top_bottom = 2πr^2.
The total surface area, A_total, is the sum of the lateral surface area and the top and bottom areas:
A_total = A_lateral + A_top_bottom
A_total = 2πrh + 2πr^2
To minimize the surface area, we can express h in terms of r from the volume equation and substitute it into the surface area equation.
10 = π * r^2 * h
h = 10 / (π * r^2)
Now substitute h in the surface area equation:
A_total = 2πrh + 2πr^2
A_total = 2πr * (10 / (π * r^2)) + 2πr^2
A_total = 20 / r + 2πr^2
To find the minimum surface area, we can take the derivative of the surface area equation with respect to r and set it equal to zero:
dA_total/dr = -20/r^2 + 4πr = 0
Simplifying the equation:
-20/r^2 + 4πr = 0
-20 + 4πr^3 = 0
4πr^3 = 20
r^3 = 20 / (4π)
r^3 = 5 / (π)
Taking the cube root of both sides:
r = (5 / (π))^(1/3)
Now that we have the value of r, we can substitute it back into the volume equation to find the corresponding value of h:
10 = π * r^2 * h
h = 10 / (π * r^2)
Substituting the value of r:
h = 10 / (π * ((5 / (π))^(1/3))^2)
Simplifying the expression:
h = 10 / (π * (5 / (π^2))^(2/3))
Now, we can simplify this further to find the numerical values of r and h.
To find the economical proportion between the radius and height of a cylindrical can that gives the least dimensions of metal to enclose a volume of 10 cubic units, we need to apply some mathematical principles.
Let's start by defining the variables we'll use:
- Let's denote the radius of the can as 'r'
- Let's denote the height of the can as 'h'
- The volume of the can is given as 10 cubic units, so we have V = 10 cu.
The volume of a cylindrical can is calculated using the formula V = πr^2h, where π is a mathematical constant approximately equal to 3.14159.
Therefore, we can rewrite our equation as 10 = πr^2h.
Now, we need to find the economical proportion, which means minimizing the amount of metal used. The amount of metal used is determined by the surface area of the can.
The surface area of a cylindrical can is calculated using the formula A = 2πrh + 2πr^2.
We can rewrite this equation in terms of the height, h: h = (A - 2πr^2)/(2πr).
Substituting this value of h into our volume equation, we get:
10 = πr^2((A - 2πr^2)/(2πr)).
Simplifying further, we can express this equation as:
20πr = A - 2πr^2.
To find the economical proportion, we need to minimize the surface area A. To do that, we take the derivative of the equation with respect to r and set it equal to zero.
Differentiating the equation, we get:
20π = -4πr.
Simplifying further, we find r = 5.
Now, substitute this value back into our volume equation to find the height, h:
10 = π(5^2)h,
10 = 25πh,
h = 10/(25π).
Therefore, for the least dimensions of metal needed to enclose a volume of 10 cubic units, the can should have a radius of 5 units and a height of 10/(25π) units.