A company wants to manufacture cylindrical aluminum cans with a value of 1000cm^3 (1 Liter) What radius and height of the can be to minimize the amount of aluminum used? Please help! This is very hard. Thanks.
You know that
pi r^2 h = 1000, so
h = 1000/(pi r^2)
the surface area is
a = 2pi r^2 + 2pi r h
= 2pi r^2 + 2pi r * 1000/(pi r^2)
= 2pi r^2 + 2000/r
da/dr = 4pi r - 2000/r^2
= (4pi r^3 - 2000)/r^2
Since r > 0, da/dr=0 when
4pi r^3 = 2000
r^3 = 2000/4pi = 500/pi
r = 5 ∛(4/pi) ~= 5.42
go to wolframalpha.com and enter
plot y = 2pi r^2 + 2000/r where 3<r<8
to see where it is minimum
To minimize the amount of aluminum used to manufacture cylindrical aluminum cans with a volume of 1000 cm³, we can use the principles of calculus and optimization.
Let's assume the radius of the can is "r" and the height is "h". We need to find the values of "r" and "h" that minimize the surface area of the cylinder.
Step 1: Find the Surface Area of the Can
The surface area of a cylinder is given by the formula:
A = 2πrh + 2πr²
Step 2: Express one variable in terms of the other
Since we know the volume of the can is 1000 cm³, we can express one variable in terms of the other using the formula for the volume of a cylinder:
V = πr²h
Solving for "h":
h = V / (πr²)
h = 1000 / (πr²)
Step 3: Substitute the expression for "h" into the surface area equation
Substitute the expression for "h" into the surface area equation:
A = 2πr(1000 / (πr²)) + 2πr²
A = 2000 / r + 2πr²
Step 4: Find the minimum surface area by differentiating with respect to "r"
Differentiate the surface area equation with respect to "r" to find the critical points:
dA/dr = -2000 / r² + 4πr
Step 5: Set the derivative equal to zero and solve for "r"
To find the critical points, set the derivative equal to zero:
-2000 / r² + 4πr = 0
-2000 + 4πr³ = 0
4πr³ = 2000
r³ = 2000 / (4π)
r³ = 500 / π
r ≈ 6.308 cm (rounded to 3 decimal places)
Step 6: Find the corresponding height "h"
Substitute the value of "r" into the expression for "h" to find the corresponding height:
h = 1000 / (πr²)
h = 1000 / (π * (6.308)²)
h ≈ 8.972 cm (rounded to 3 decimal places)
Therefore, to minimize the amount of aluminum used, the radius of the can should be approximately 6.308 cm and the height should be approximately 8.972 cm.
To minimize the amount of aluminum used in the manufacturing of cylindrical aluminum cans with a volume of 1000 cm³, you can use calculus to find the dimensions that optimize the design.
Let's start by determining the mathematical relationship between the radius and height of the can.
We know that the volume of a cylinder is given by the formula:
V = πr²h
We are given that the volume, V, is 1000 cm³. Rearranging the formula, we get:
h = 1000 / (πr²)
Now, we can express the amount of aluminum used in terms of the radius (r) and height (h) of the can. The amount of aluminum used is directly proportional to the surface area of the can, which is given by:
A = 2πr² + 2πrh
We can now substitute the expression for height (h) into the area equation:
A = 2πr² + 2πr(1000 / (πr²))
A = 2πr² + 2000 / r
To find the dimensions that minimize the amount of aluminum used, we need to differentiate the area equation with respect to the radius (r) and set the derivative equal to zero. This will give us the critical points, one of which will correspond to the minimum.
dA/dr = 4πr - 2000 / r² = 0
Now we can solve for r:
4πr = 2000 / r²
r³ = 500 / π
r = (500 / π)^(1/3)
Now that we have the radius (r), we can substitute it back into the expression for height (h) to find the corresponding height:
h = 1000 / (πr²)
h = 1000 / (π * [(500 / π)^(1/3)]²)
Simplifying this equation will give you the value of h. Plug it back into the volume equation to ensure the volume remains 1000 cm³.
Please note that I have provided the mathematical steps to find the values of r and h that minimize the amount of aluminum used. You can use a calculator or a math software to calculate the numerical values.