A manufacturer of aluminum cans has budgeted 20ð square centimeters of material for each can. What is the maximum volume that such a can could hold?
Radius = r
Height = h
Volume, V = πr²h
Surface area, S = 2πr²+ 2πrh
=2πr(r+h)
Given S=20π, therefore
2πr(r+h) = 20π, or
r(r+h)=10
h=(10/r-r)
V(r) = πr²h
=πr²(10/r-r)
=π(10r-r³)
Since V(r) is now expressed in terms of r, calculate the derivative V'(r).
Equating V'(r)=0 gives the value of r that maximizes or minimizes the volume.
I get r=1.8 approx.
Calculate the second derivative V"(r) of the volume with respect to r, and evaluate V"(1.8) to make sure it is negative, which implies V(1.8) is a maximum.
I get Vmax=38 approx.
Check my work and my thinking.
let the radius of the can be r cm and its height be h cm
2pi(r^2) + 2pi(rh) = 20ð
(I don't know what the symbol at the end of 20ð is, I will continue assuming it is 200. If it is anything else, you will have to adjust the calculations)
2pi(r^2) + 2pi(rh) = 200
h = 100/(pir) - r
volume = pi(r^2)h
= pir^2(100/(pir) - r)
= 100r - pr^3
d(volume)/dr = 100 - 2pir^2 = 0 for a max/min of volume
2pir^2 = 100
r^2 = 50/pi
r = 3.989
then h = 3.989
which confirms the concept that the largest volume of a can using the least material is obtained when the height and radius are the same.
To determine the maximum volume that the can could hold, we need to relate the given budgeted material area to the volume of the can.
The volume of a cylindrical can is calculated using the formula V = πr^2h, where V is the volume, π is a constant (~3.14159), r is the radius of the can's base, and h is the height of the can.
To solve the problem, we can start by expressing the given budgeted material area in terms of the equation for the surface area of the can. Since the surface area of a cylinder is given by A = 2πrh + 2πr^2, where A is the surface area, we have:
20π = 2πrh + 2πr^2
Simplifying the equation, we get:
10π = πrh + πr^2
Dividing by π, we have:
10 = rh + r^2
Now, let's solve the equation for h:
h = (10 - r^2) / r
Substitute this expression for h into the volume formula:
V = πr^2((10 - r^2) / r)
V = πr(10 - r^2)
To find the maximum volume, we can take the derivative of V with respect to r, set it equal to zero, and solve for r. Let's calculate the derivative:
dV/dr = π(10 - 3r^2)
Setting dV/dr equal to zero:
10 - 3r^2 = 0
3r^2 = 10
r^2 = 10/3
r ≈ √(10/3)
Now, substitute this value of r back into the equation for h:
h = (10 - (√(10/3))^2) / (√(10/3))
h ≈ (10 - 10/3) / (√(10/3))
Simplify:
h ≈ 20/(3√10)
Finally, compute the maximum volume V:
V = π(√(10/3))^2(20/(3√10))
V ≈ (10/3)(20/(3√10))^2π
V ≈ 200/(27√10)π
Therefore, the maximum volume that the can can hold is approximately 200/(27√10)π cubic centimeters.