A cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides
of the can costs 0.02 cents per square centimeter. The material for the top and bottom of the can need to be thicker,
and costs 0.05 cents per square centimeter. Find the dimensions for the can that will minimize production cost.
okay. one more time
πr^2h = 200
h = 200/(πr^2)
The area is thus
2πrh + 2πr^2
That makes the cost c
c = .02*2πr(200/πr^2) + .05*2πr^2
c = 4/r + 0.1πr^2
So to find minimum c, set dc/dr = 0
That occurs when r = ∛(20/π) = 1.853
Now finish it off
To find the dimensions that will minimize production cost, we can set up a mathematical model and use calculus to find the optimal solution.
Let's denote the radius of the cylinder as r and the height as h.
The volume of a cylinder is given by the formula V = πr^2h. In this case, we have V = 200 cubic centimeters.
To minimize the production cost, we need to minimize the cost of the side material and the cost of the top and bottom material.
The cost of the side material (C1) is given by the formula C1 = 0.02 * (2πrh).
The cost of the top and bottom material (C2) is given by the formula C2 = 0.05 * (2πr^2).
The total production cost (C) is the sum of C1 and C2: C = C1 + C2.
Now, let's substitute the formulas for C1 and C2 into C and simplify it:
C = 0.02 * (2πrh) + 0.05 * (2πr^2)
= 0.04πrh + 0.10πr^2
= π(0.04rh + 0.10r^2)
To find the dimensions that minimize the production cost, we need to find the values of r and h that minimize C.
Next, we'll need to express h in terms of r using the volume formula V.
V = πr^2h
200 = πr^2h
h = 200 / (πr^2)
Now, substitute this expression for h into the cost formula C:
C = π(0.04r(200 / (πr^2)) + 0.10r^2)
= 0.04r(200 / r^2) + 0.10r^2
= 8 / r + 0.10r^2
To minimize C, we need to find the critical points by taking the derivative of C with respect to r and setting it equal to zero.
dC/dr = -8 / r^2 + 0.20r
0 = -8 / r^2 + 0.20r
Multiply the equation by r^2 to eliminate the fractions:
0 = -8 + 0.20r^3
8 = 0.20r^3
r^3 = 8 / 0.20
r^3 = 40
r = (40)^(1/3)
Now, substitute this value of r back into the equation for h:
h = 200 / (πr^2)
h = 200 / (π(40)^(2/3))
So, the dimensions that will minimize production cost are:
- The radius, r = (40)^(1/3) cubic centimeters.
- The height, h = 200 / (π(40)^(2/3)) cubic centimeters.
Please note that the values of r and h obtained here are approximate, as we rounded pi to its decimal approximation.
To minimize the production cost of the can, we need to find the dimensions that minimize the surface area of the can.
Let's start by defining the variables:
Let r be the radius of the top and bottom of the can.
Let h be the height of the can.
Since the can is in the shape of a cylinder, the volume of the can can be calculated using the formula for the volume of a cylinder:
V = πr^2h
We are given that the volume needs to be 200 cubic centimeters, so we can write the equation as:
200 = πr^2h
To minimize the surface area, we need to consider the cost of the material for the sides and the cost of the material for the top and bottom.
The cost of the material for the sides is 0.02 cents per square centimeter, and the cost of the material for the top and bottom is 0.05 cents per square centimeter.
The surface area of the sides of the can is given by the formula:
A_sides = 2πrh
The surface area of the top and bottom of the can is given by the formula:
A_top_bottom = 2πr^2
The total cost of materials can be calculated as:
Cost = 0.02A_sides + 0.05A_top_bottom
Substituting the formulas for the surface area into the cost equation, we get:
Cost = 0.02(2πrh) + 0.05(2πr^2)
Simplifying the equation, we have:
Cost = 0.04πrh + 0.1πr^2
To minimize the cost, we need to take the partial derivatives of Cost with respect to r and h, and set them equal to zero.
dCost/dr = 0.04πh + 0.2πr = 0
dCost/dh = 0.04πr = 0
Solving these equations, we find that r = 0 and h = 0, which are not meaningful solutions.
Instead, we can use the volume equation to solve for one variable in terms of the other and substitute it into the cost equation.
From the volume equation, we can solve for h:
h = 200/(πr^2)
Substituting this expression for h into the cost equation, we have:
Cost = 0.04πr(200/(πr^2)) + 0.1πr^2
Simplifying the equation, we get:
Cost = 8/r + 0.1πr^2
To find the minimum cost, we take the derivative of Cost with respect to r and set it equal to zero:
dCost/dr = -8/r^2 + 0.2πr = 0
Solving for r, we find:
-8/r^2 + 0.2πr = 0
Multiplying through by r^2, we get:
-8 + 0.2πr^3 = 0
Solving for r, we find:
r^3 = 40/(π)
Taking the cube root of both sides, we get:
r = (40/(π))^(1/3)
Substituting this value of r into the volume equation, we find:
h = 200/(π((40/(π))^(1/3))^2)
Simplifying the equation, we have:
h = 200/(40^(2/3)/π^(1/3))
h = (200π^(1/3))/40^(2/3)
Therefore, the dimensions for the can that will minimize production cost are:
radius, r = (40/(π))^(1/3)
height, h = (200π^(1/3))/40^(2/3)