A company needs to make a cylindrical can that can hold precisely 2.1 liters of liquid. If the entire can is to be made out of the same material, find the dimensions of the can that will minimize the cost. Round your answer to the nearest four decimal places.
Why did the can go to therapy? It had trouble finding its dimensions!
To find the optimal dimensions, we'll use a little math humor. Let's say the radius of the can is 'r' and the height is 'h'. The volume of a cylinder is given by V = πr^2h. We want to find the dimensions that minimize the cost, which is directly proportional to the surface area of the can.
The surface area of a cylinder is given by A = 2πrh + 2πr^2. Since we want to minimize the cost, we need to minimize the surface area.
To continue with the math fun, let's solve for one variable (r or h) in one equation and substitute it into the other equation. We can solve for h in terms of r by using the formula for volume: h = V / (πr^2).
Substituting h into the equation for surface area, we get A = 2πr(V / (πr^2)) + 2πr^2 which simplifies to A = 2V/r + 2πr^2.
To minimize the surface area, we differentiate with respect to r: dA/dr = -2V/r^2 + 4πr.
Setting the derivative equal to zero (dA/dr = 0), we get -2V/r^2 + 4πr = 0.
Solving for r, we find r = (V / (2π))^1/3.
Now that we have r, we can substitute it back into the equation for h to find h = 2V / (πr^2).
Finally, we can plug in the given volume V = 2.1 liters (or 2100 cubic centimeters) to calculate the dimensions of the can that minimize the cost.
r = (2100 / (2π))^1/3 ≈ 6.1412 cm (rounded to four decimal places).
h = 2 * 2100 / (π * (6.1412)^2) ≈ 17.4228 cm (rounded to four decimal places).
So, the optimal dimensions for the can that minimize the cost are approximately a radius of 6.1412 cm and a height of 17.4228 cm. Just remember, this answer may not fit all your specific requirements, so it's always a good idea to consult with engineers or professionals for precise calculations.
To find the dimensions of the can that will minimize the cost, we need to consider the cost function. Let's assume the cost of material per unit area is $C per square unit.
The first step is to determine the mathematical relationship between the volume of the cylinder, the cost, and the dimensions. To do this, we can set up the following equation:
Cost = (Cost per unit area) × (Total surface area of the cylindrical can)
The total surface area of the cylindrical can is the sum of the curved surface area (2πrh) and the area of the two circular bases (2πr^2).
Now let's proceed to the step-by-step solution:
Step 1: Write the equation for the cost function.
Cost = (Cost per unit area) × (2πrh + 2πr^2)
Step 2: Express the height (h) in terms of the radius (r) and the volume (V).
The volume of a cylinder is given by V = πr^2h. Rearranging this equation to solve for h, we have:
h = V / (πr^2)
Step 3: Plug in the expression for h in the cost function.
Cost = (Cost per unit area) × (2πr(V / (πr^2)) + 2πr^2)
Step 4: Simplify the cost function.
Cost = 2π(V/r + r^2) × (Cost per unit area)
Step 5: Differentiate the cost function with respect to r to find the critical points.
d(Cost)/dr = 0
Step 6: Solve for the value of r that minimizes the cost by setting the derivative equal to 0 and solving for r.
2π(V/r + r^2) × (Cost per unit area) = 0
Step 7: Solve for r. Rearrange the equation to get:
V/r + r^2 = 0
V + r^3 = 0
r^3 = -V
Step 8: Solve for r by taking the cube root of both sides.
r = (-V)^(1/3)
Step 9: Substitute the value of r back into the expression for h to find the corresponding height.
h = V / (πr^2)
Step 10: Substitute the given value of V (2.1 liters) into the equations for r and h to find their respective values.
r ≈ (-2.1)^(1/3) ≈ -1.289
h ≈ 2.1 / (π(-1.289)^2) ≈ 0.820
Since negative dimensions don't make sense in this context, we take the absolute values of the dimensions.
The dimensions of the can that will minimize the cost (rounded to four decimal places) are:
radius (r) ≈ 1.289
height (h) ≈ 0.820
To find the dimensions of the can that will minimize the cost, we need to consider the cost function and optimize it.
Let's define the variables:
r: the radius of the cylindrical can
h: the height of the cylindrical can
We need to find the values of r and h that minimize the cost of manufacturing the can.
First, let's derive a formula for the cost of manufacturing the can.
The cost of the material needed to make the can is directly proportional to its surface area.
The surface area of the can consists of two parts:
1. The curved surface area (excluding the top and bottom surfaces) which can be calculated using the formula for the lateral surface area of a cylinder: 2πrh.
2. The area of the two circular ends of the can, which can be calculated using the formula for the area of a circle: 2πr^2.
Since we want to minimize the cost, we need to minimize the surface area of the can.
Now, let's set up the equation for the volume of the can.
The volume V of a cylinder can be calculated using the formula: V = πr^2h.
Given that the can needs to hold precisely 2.1 liters of liquid, which is equivalent to 2100 cubic centimeters, we can write the equation:
2100 = πr^2h.
Now, let's solve this equation for r in terms of h to eliminate one variable.
We can rearrange the equation as follows:
r^2 = (2100 / (πh))
Taking the square root of both sides, we get:
r = √(2100 / (πh))
Next, substitute this expression for r into the equation for the surface area of the can:
A = 2πrh + 2πr^2
Now, we can express the cost C as a function of h:
C = k(2πrh + 2πr^2)
Here, k represents the cost per unit surface area.
To find the dimensions that minimize the cost, we need to find the value of h that minimizes the cost function C(h). We can achieve this by applying the concept of calculus and finding the critical points.
Take the derivative of the cost function with respect to h, set it equal to zero, and solve for h to find the critical point(s).
Once we find the critical point(s), plug the value(s) into the expression for r to obtain the corresponding radius value(s).
Finally, substitute the values of r and h back into the cost function to calculate the minimum cost.
Rounding the answer to the nearest four decimal places will give us the dimensions of the can that minimize the cost.
area = A = 2 pi r h + 2 pi r^2
2.1 liters = .0021 m^3
.0021 = pi r^2 h
so
h = .0021/(pi r^2)
so
A = 2 pi r (.0021)/(pi r^2) + 2 pi r^2
or
A = .0042/r + 2 pi r^2
for min area dA/dr = 0
0 = -.0042/r^2 + 4 pi r
4 pi r^3 = .0042
r = .0694 meters or 6.94 cm
then h = .139 meters or 13.9 cm