A cylinder shaped can needs to be constructed to hold 300 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.07 cents per square centimeter. Find the dimensions for the can that will minimize production cost.
Helpful information:
h : height of can, r : radius of can
Volume of a cylinder: V=πr^2h
Area of the sides: A=2πrh
Area of the top/bottom: A=πr^2
To minimize the cost of the can:
Radius of the can: ___
Height of the can: ___
Minimum cost: ___cents
cost = .07(top + bottom) + .04(side)
= .07(2πr^2) + .04(2π r h)
now we know that πr^2 h = 300
h = 300/(πr^2)
cost = .14π r^2 + .08π r (300/πr^2)
= .14π r^2 + 24/r
d cost / dr = .28π r - 24/r^2 = 0 for a min of cost
.28π r^3 = 24
r^3 = 24 / .28π = 27.2837....
r = 27.2837....^(1/3) = 3.04047...
h = 10.5366...
round to whatever accuracy you need
plug r = .... into the cost equation
cost = .14π r^2 + 24/r
To minimize the cost of the can, we need to optimize the dimensions of the can based on the given cost of materials.
Let's start by expressing the cost of the can in terms of its dimensions.
The cost of the side material is given as 0.04 cents per square centimeter, and the cost of the top and bottom material is given as 0.07 cents per square centimeter.
The cost of the side material (Cside) can be calculated as follows:
Cside = 0.04 * Area of the sides
The cost of the top and bottom material (Ctop_bottom) can be calculated as follows:
Ctop_bottom = 0.07 * 2 * Area of the top/bottom (each top and bottom)
The total cost (Ctotal) can be calculated as follows:
Ctotal = Cside + Ctop_bottom
Now, let's substitute the respective formulas for the area and simplify the cost equation.
Cside = 0.04 * 2πrh
Ctop_bottom = 0.07 * 2 * πr^2
Ctotal = 0.08πrh + 0.14πr^2
Since we are given that the volume of the can needs to be 300 cubic centimeters, we can express the height (h) in terms of the radius (r) using the volume formula for a cylinder:
V = πr^2h
300 = πr^2h
h = 300/πr^2
Now, we can substitute the expression for h into the cost equation and simplify:
Ctotal = 0.08πr(300/πr^2) + 0.14πr^2
Ctotal = 24/r + 0.14πr^2
To minimize the cost, we differentiate the cost equation with respect to r and find the value of r that makes the derivative equal to zero:
dCtotal/dr = -24/r^2 + 0.28πr = 0
Multiplying through by r^2 to clear the fraction:
-24 + 0.28πr^3 = 0
Solving for r:
0.28πr^3 = 24
r^3 = 24/(0.28π)
r^3 = 27/(0.28π)
r = (27/(0.28π))^(1/3)
We can now substitute the value of r back into the expression for h to find the value of h:
h = 300/π((27/(0.28π))^(1/3))^2
Finally, we can substitute the values of r and h into the cost equation to find the minimum cost:
Ctotal = 24/((27/(0.28π))^(1/3)) + 0.14π((27/(0.28π))^(1/3))^2
Calculating the values, we get:
Radius of the can: approximately 1.729 cm
Height of the can: approximately 13.799 cm
Minimum cost: approximately 325.088 cents
To minimize the cost of the can, we need to consider the cost of both the sides and the top/bottom of the can.
Let's start by finding an equation for the cost. The cost of the sides will be the cost per square centimeter multiplied by the area of the sides, while the cost of the top/bottom will be the cost per square centimeter multiplied by the area of the top/bottom.
The equation for the cost C is:
C = (0.04 * A_sides) + (0.07 * A_top/bottom)
Now, let's express the areas in terms of the radius (r) and height (h) of the can.
The area of the sides A_sides is given by:
A_sides = 2πrh
The area of the top/bottom A_top/bottom is given by:
A_top/bottom = 2πr^2
Since we are trying to minimize the cost, we can substitute these equations back into the cost equation to get a function of just r and h.
C = (0.04 * 2πrh) + (0.07 * 2πr^2)
C = 0.08πrh + 0.14πr^2
Now we can differentiate C with respect to r and h, and find the critical points (where the partial derivatives are equal to zero) to determine the dimensions that minimize the cost.
∂C/∂r = 0.08πh + 0.28πr = 0
∂C/∂h = 0.08πr = 0
From the second equation, we can see that r = 0. From the first equation, we can solve for h:
0.08πh + 0.28πr = 0
0.08πh + 0.28π(0) = 0
0.08πh = 0
Since h cannot be zero, there is no critical point in this case.
However, we can observe that the cost C will increase as the dimensions (r and h) of the can increase without bounds. Therefore, to minimize the cost of the can, we should consider the dimensions that are practical and reasonable for the problem.
For example, we can set a reasonable value for the height (h) and calculate the radius (r) that corresponds to a volume of 300 cm^3.
Using the volume equation V = πr^2h, we can solve for r:
300 = πr^2h
r^2 = 300 / (πh)
r = √(300 / (πh))
Now we have the relation between r and h for a given volume. We can substitute this value of r into the cost equation C = 0.08πrh + 0.14πr^2, using the chosen value for h, and calculate the minimum cost.
For example, let's say we choose h = 10 cm:
r = √(300 / (π * 10))
r ≈ 3.271 cm
Now we can substitute this value of r into the cost equation:
C = (0.04 * 2π * 3.271 * 10) + (0.07 * 2π * 3.271^2)
C ≈ 8.95 cents
Therefore, for a can with a height of 10 cm, the radius that minimizes the cost is approximately 3.271 cm, and the minimum cost is approximately 8.95 cents.