A cylindrical can is to hold 20 pie mraised to the 3rd power. The material for the top and bottom costs $10/m raised to the 2nd power and material for the side costs $8/m to the 2nd power. Find the radius r and height h of the most economical can.
Volume = pi r^2 h = 20 pi (m^3)
Therefore r^2 h = 20
h = 20/r^2
Cost = 2 pi r^2 * 10 + 2 pi r h * 8
= 20 pi r^2 + 16 pi r(20/r^2)
= 20 pi r^2 + 320 pi/r
Set the derivative with respect to r equal to zero.
40 pi r = 320 pi r^-2
r^3 = 80
r = 4.31 m
Check my math
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 (2πr^2), and the cost of the side is $8 times the surface area of the side (2πrh).
So, the total cost C is given by:
C = 20($10)(2πr^2) + 20($8)(2πrh).
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)(4πr) + 20($8)(2πh) = 0.
dC/dh = 20($8)(2πr) = 0.
Simplifying, we have:
40πr + 80πh = 0,
16πr = 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 20π^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/(πr^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)(2πr^2) + 20($8)(2πr)(20/(πr^2))
= 400πr^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 = 800πr - 3200/r^2 = 0.
Multiplying through by r^2, we get:
800πr^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/(πr^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!
To find the dimensions of the most economical can, we need to minimize the cost of the materials used.
Let's denote the radius of the can as "r" and the height of the can as "h". The volume of the can is given as 20π cubic meters.
Volume of a cylinder = π * r^2 * h
Given: π * r^2 * h = 20π
Simplifying the equation, we have:
r^2 * h = 20
Now, let's determine the cost of materials used for the can.
Cost of the top and bottom material = 2 * Cost per unit area * Area of a circle
Cost of the side material = Cost per unit area * Area of the curved surface
Given: Cost per unit area for top and bottom = $10/m^2
Cost per unit area for side = $8/m^2
Area of a circle = π * r^2
Area of the curved surface = 2π * r * h
Cost of materials used = Cost of top and bottom material + Cost of side material
= 2 * (10 * π * r^2) + 8 * (2 * π * r * h)
= 20πr^2 + 16πrh
Now we need to minimize the cost function, so let's differentiate it with respect to "r" and "h" to find the critical points.
d(Cost)/dr = 40πr + 16πh
d(Cost)/dh = 16πr
Setting both derivatives equal to zero, we have:
40πr + 16πh = 0 ...(1)
16πr = 0 ...(2)
From equation (2), r = 0, which is not possible for the dimensions of a can. So, we can ignore it.
Substituting equation (2) into equation (1), we have:
40π(20/r^2) + 16πh = 0
8000/r^2 + 16πh = 0
h = -8000/(16πr^2)
h = -500/r^2
Since both "r" and "h" cannot be negative, we can ignore the negative solution.
Now let's substitute the value of "h" from equation (1) into the volume equation:
r^2 * h = 20
r^2 * (-500/r^2) = 20
-500 = 20
This equation has no valid solution, which means we made a mistake somewhere.
Please recheck the problem statement and ensure all the given information is accurate.
To find the dimensions of the most economical can, we need to minimize the cost of the materials used. Let's break down the problem into smaller steps:
Step 1: Determine the volume of the cylindrical can.
The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height of the cylinder. In this case, we know the volume is 20π (since 20π is equivalent to 20π * 1^2, the base area of the cylinder). So, we can write the equation: 20π = πr^2h.
Step 2: Determine the cost of the materials for the top and bottom.
The cost of the top and bottom of the can is given as $10/m^2. Since the top and bottom of the can form two circles, which have an area of πr^2 each, the total cost for the top and bottom is 2 * 10πr^2 dollars.
Step 3: Determine the cost of the material for the side.
The side of the can is a rectangle that wraps around the circumference of the base. The area of the side is given by A = 2πrh, as there are two sides on the cylinder. The cost of the material for the side is $8/m^2, so the total cost is 8 * 2πrh dollars.
Step 4: Determine the total cost of the can.
To find the total cost, we sum up the cost of the top and bottom (Step 2) and the cost of the side (Step 3). So, the total cost of the can, C, is given by: C = 2 * 10πr^2 + 8 * 2πrh.
Step 5: Express C in terms of h only.
To find the radius and height that minimize the cost, we need to express the total cost C in terms of h only. We can substitute the value of r from the equation in Step 1 into the equation for C. This gives us: C = 2 * 10π(20π/h^2) + 8 * 2πh(20/h).
Step 6: Simplify the equation for C.
Simplify the equation by canceling common factors and combining terms where possible. After simplification, the equation becomes: C = 400π^2/h + 320π.
Step 7: Minimize the cost function.
To find the value of h that minimizes the cost, we take the derivative of C with respect to h and set it equal to zero. Then solve for h.
dC/dh = -400π^2/h^2 + 0.
Solving for h, we get h = √(400π^2/400π) = √π.
Step 8: Find the corresponding value of r.
Using the value of h obtained in Step 7, we can substitute it back into the equation in Step 1 to find the corresponding value of r. Plugging h = √π into the equation, we get:
20π = πr^2(√π),
which simplifies to r = √(20/π).
Therefore, the dimensions of the most economical can are r = √(20/π) and h = √π.