A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs 0.03 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.06 cents per square centimeter. Find the dimensions for the can that will minimize production cost.
let the radius be r
let the height be h
Volume = πr^2h
πr^2h = 400
h = 400/(πr^2)
cost = different prices x surface areas
= .03(2πrh) + 2(.06) πr^2
= .03[2πr(400/πr^2) + 4πr^2]
= .03[ 800/r + 4πr^2]
d(cost)/dr = .03[ -800/r^2 + 8πr] = 0 for a min of cost
800/r^2 = 8πr
100/π = r^3
r = 3.169
h = 400/(π(3.169)^2) = 12.679
check my arithmetic
To minimize the production cost, we'll need to find the dimensions (radius and height) of the cylinder that minimize the surface area while holding 400 cubic centimeters of soup.
Let's denote the radius of the cylinder as "r" and the height as "h." We can start by writing the equation for the volume of the cylinder:
Volume = πr^2h = 400 cm^3 (1)
To find the surface area of the cylinder, we need to calculate the area of the sides, top, and bottom separately. The area of the sides can be calculated as follows:
Area of sides = 2πrh
The area of the top and bottom can be calculated using the formula for the area of a circle:
Area of top or bottom = 2πr^2
Now, the total surface area (A) of the can will be the sum of the areas of the sides, top, and bottom:
A = 2πrh + 2πr^2 + 2πr^2
A = 2πrh + 4πr^2 (2)
Now, we can substitute the value of πr^2h from equation (1) into equation (2):
A = 2πrh + 4(400/π)
A = 2πrh + 1600/π
To minimize the production cost, we'll need to find the minimum of the cost function, which is directly proportional to the surface area (A). The cost function (C) can be given as:
C = (cost per square centimeter of sides) * (area of sides) + (cost per square centimeter of top/bottom) * (area of top/bottom)
C = 0.03 * (2πrh) + 0.06 * (2πr^2)
Now let's simplify the equation for the total cost of production:
C = 0.06πr(2h + 2r) + 0.06 * 1600/π
C = 0.06πr(2h + 2r) + 96/π
To find the dimensions that minimize the cost, we'll need to differentiate the cost function with respect to both r and h, and then solve the resulting equations to find the critical points.
Differentiating with respect to r:
dC/dr = 0.06π(2h + 2r) - 0.06πr = 0
0.06π(2h + 2r) = 0.06πr
2h + 2r = r
2h = r
Differentiating with respect to h:
dC/dh = 0.06πr(2) = 0.12πr = 0
From the second equation, 0.12πr = 0, we get r = 0, which is not a valid solution.
Substituting r = 2h into the equation 2h + 2r = r, we get:
2h + 2(2h) = 2h
6h = 2h
4h = 0
Again, this gives an invalid solution. Hence, there are no critical points.
However, we know that the cost is a continuous function, so we also need to check the endpoints of the feasible region, which satisfy the volume equation (1).
Let's solve the volume equation for h:
πr^2h = 400
h = 400 / (πr^2)
Substituting this value of h into our cost equation C:
C(r) = 0.06πr(2h + 2r) + 96/π
C(r) = 0.06πr(2(400/(πr^2)) + 2r) + 96/π
C(r) = 0.12(400/r^2) + 0.12πr + 96/π
Simplifying further:
C(r) = 4800/r^2 + 0.12πr + 96/π
To find the minimum, we can take the derivative with respect to r and set it equal to zero:
dC/dr = -9600/r^3 + 0.12π = 0
9600/r^3 = 0.12π
Solving for r^3:
r^3 = 9600 / (0.12π)
r^3 ≈ 21739.13
r ≈ 26.79 cm
Substituting this value back into the volume equation (1) to solve for h:
π(26.79^2)h = 400
h ≈ 0.56 cm
Therefore, the dimensions that will minimize the production cost are approximately:
radius (r) ≈ 26.79 cm
height (h) ≈ 0.56 cm
To find the dimensions that will minimize the production cost of the can, we can use optimization techniques.
Let's assume that the can has a radius of "r" centimeters and a height of "h" centimeters. We want to minimize the cost, which is the sum of the costs for the sides, top, and bottom of the can.
1. The volume of the cylinder is given by the formula: V = πr^2h. We know that the volume of the can should be 400 cubic centimeters. So we have the equation: πr^2h = 400.
2. The cost for the sides of the can is given by the formula: Cs = 0.03πrh. This is because the sides of the can are a cylinder without the top and bottom.
3. The cost for the top and bottom of the can is given by the formula: Ctb = 2(0.06πr^2). This is because the top and bottom of the can are both circles with radius "r".
Now, we need to express one variable in terms of the other. Let's solve the volume equation for "h": h = 400 / (πr^2).
Substituting this value of "h" in the cost equation for the sides, we get: Cs = 0.03πr(400 / (πr^2)) = 12/r.
Substituting the same value of "h" in the cost equation for the top and bottom, we get: Ctb = 2(0.06πr^2) = 0.12πr^2.
The total cost (C) is the sum of the cost for the sides and the cost for the top and bottom: C = Cs + Ctb = 12/r + 0.12πr^2.
To find the dimensions of the can that minimize the cost, we need to find the minimum value of C with respect to "r". We can do this by taking the derivative of C with respect to "r" and setting it equal to zero.
dC/dr = -12/r^2 + 0.24πr = 0.
Simplifying the above equation, we get: 0.24πr = 12/r^2.
Multiplying both sides by r^2, we obtain: 0.24πr^3 = 12.
Dividing both sides by 0.24π, we find: r^3 = 50 / π.
Taking the cube root of both sides yields: r ≈ (50 / π)^(1/3) ≈ 2.227 cm (rounded to three decimal places).
Now that we have the value of "r," we can substitute it back into the equation for "h": h = 400 / (πr^2) ≈ 4.496 cm (rounded to three decimal places).
Therefore, the dimensions of the can that will minimize production cost are approximately: radius ≈ 2.227 cm and height ≈ 4.496 cm.