A company needs to make a cylindrical can that can hold precisely 1.5 liters of liquid. If the entire can is to be made out of the same material, find the dimensions (radius and height) of the can that will minimize the cost. Round your answer to the nearest four decimal places.
Recall that the volume of a cylinder is πr^2h and the surface area is 2πrh+2πr^2 where r is the radius and h is the height. Also note that 1 liter is equal to 1000cm3.
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wheres the answer?
To solve this problem, we need to find the dimensions (radius and height) of the can that minimize the cost. The cost is directly related to the surface area of the can, since the company needs to use a material that covers the entire can.
Step 1: Convert the volume from liters to cubic centimeters (cm^3)
Since 1 liter is equal to 1000 cm^3, 1.5 liters would be equal to 1500 cm^3.
Step 2: Write down the constraints
The volume of the cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height. Therefore, we have the constraint V = πr^2h = 1500 cm^3.
Step 3: Write down the cost function
The cost of the can is related to the surface area, which is given by the formula S = 2πrh + 2πr^2, where S is the surface area. Therefore, the cost function C can be written as C = k(2πrh + 2πr^2), where k is a constant representing the cost per unit area. Since k is constant, we can ignore it for the purpose of minimizing the cost function.
Step 4: Solve the constraint for one of the variables
From the constraint equation, we can solve for h in terms of r: h = 1500 / (πr^2).
Step 5: Substitute the expression for h into the cost function
Substituting the expression for h into the cost function, we have C = k(2πr(1500 / (πr^2)) + 2πr^2) = k(3000 / r + 2πr^2).
Step 6: Minimize the cost function using calculus
To minimize the cost function, we need to differentiate it with respect to r, and set the derivative equal to zero. Let's differentiate C with respect to r:
dC/dr = 3000k / r^2 + 4πkr.
Setting the derivative equal to zero, we have:
3000k / r^2 + 4πkr = 0.
Simplifying the equation:
3000k = -4πkr^3.
Dividing both sides by -4πkr^3, we get:
-750k / (πr^2) = r.
Step 7: Solve for r
Simplifying the equation further:
r^3 = -750k / (πr).
Multiplying both sides by r:
r^4 = -750k / π.
Taking the fourth root of both sides:
r = (sqrt[4](-750k / π)).
Step 8: Calculate the radius and height values
Using the given information, we can substitute k = 1 (since the constant can be ignored) and π ≈ 3.1416 into the equation to get the approximate values for the radius and height.
r ≈ (sqrt[4](-750 / 3.1416))
≈ 6.6541 (rounded to four decimal places).
Using the constraint equation, we can find the value for h:
h = 1500 / (π(6.6541)^2)
≈ 11.3274 (rounded to four decimal places).
Therefore, the dimensions (radius and height) of the can that will minimize the cost are approximately r ≈ 6.6541 cm and h ≈ 11.3274 cm.