Well, let's solve this problem together, shall we?
First, let's start by finding the general formula for the surface area of the cylindrical can. The formula for the surface area is given by:
A = 2πrh + πr^2.
Since we want to find the most economical can, we need to minimize the cost. The cost of the top and bottom is $10 times the surface area of the top or bottom (2Ï€r^2), and the cost of the side is $8 times the surface area of the side (2Ï€rh).
So, the total cost C is given by:
C = 20($10)(2Ï€r^2) + 20($8)(2Ï€rh).
Now, we have two variables, r (radius) and h (height), but only one equation. Let's use some calculus to solve this problem.
We'll differentiate the cost function C with respect to r and h, and set them equal to zero to find the critical points.
dC/dr = 20($10)(4Ï€r) + 20($8)(2Ï€h) = 0.
dC/dh = 20($8)(2Ï€r) = 0.
Simplifying, we have:
40Ï€r + 80Ï€h = 0,
16Ï€r = 0.
From the second equation, we find that r = 0. Uh-oh, we can't have a can with a radius of zero. That's just a fictional can or maybe a superhero's secret weapon!
Therefore, we can conclude that there are no critical points for the cost function C. This means that there is no minimum or maximum, and we have to evaluate the endpoints.
Now, since the volume of the can is fixed at 20Ï€^3, we can use the formula for the volume of a cylinder:
V = πr^2h = 20π.
Let's solve for h in terms of r:
h = 20/(Ï€r^2).
Substituting this value of h in the equation for C, we can now write the cost function as a function of r only (since we don't need h anymore):
C = 20($10)(2Ï€r^2) + 20($8)(2Ï€r)(20/(Ï€r^2))
= 400Ï€r^2 + 3200/r.
To find the minimum cost, we can now differentiate this function with respect to r and set it equal to zero:
dC/dr = 800Ï€r - 3200/r^2 = 0.
Multiplying through by r^2, we get:
800Ï€r^3 - 3200 = 0.
Simplifying, we have:
r^3 = 4,
r = ∛4.
Using a calculator, we find that r ≈ 1.5874.
Now, we can substitute this value of r back into the equation for h:
h = 20/(Ï€r^2)
= 20/(π(∛4)^2)
= 5/(π∛4).
Using a calculator to compute, we find that h ≈ 2.0064.
So, the approximate dimensions of the most economical can are:
Radius (r) ≈ 1.5874 units,
Height (h) ≈ 2.0064 units.
Just remember, these are only approximations. And remember, when it comes to can sizes, it's not just about the size, but also about what's inside!