A cylinder shaped can needs to be constructed to hold 250 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 0.05cents per square centimeter. Find the dimensions for the can that will minimize production cost
i) Radius of the can.
ii) Height of the can.
iii) Minimum cost
Complete solution pls
Can you solve i-iii for me? I dont know how to solve it
Oh, it's soup time! Let's find the best dimensions for our can to minimize costs.
First, let's determine the surface area of the can. The can has a cylindrical shape with two circular tops, so the total surface area is given by:
Area = (2 * circular top area) + (cylindrical side area)
The circular top area is the same for both tops, so we need to calculate it only once. The cylindrical side area depends on the radius and height of the can, so we'll calculate it separately.
i) Radius of the can:
Let's say the radius is 'r'. The circular top area is given by the formula for the area of a circle:
Top Area = π * r²
ii) Height of the can:
Let's say the height is 'h'. The cylindrical side area is given by the formula for the lateral surface area of a cylinder:
Side Area = 2 * π * r * h
The volume of the can is given as 250 cubic centimeters, and we know that V = π * r² * h. We can use this equation to find the relationship between the radius and height:
r² * h = 250 / π
Solving for h, we get:
h = (250 / (π * r²))
Now, let's plug in this value of h in the equation for the side area:
Side Area = 2 * π * r * (250 / (π * r²))
= 500 / r
Now we know the formulas for the top area and side area in terms of the radius 'r'. The total surface area of the can is:
Area = (2 * Top Area) + Side Area
= (2 * π * r²) + (500 / r)
Finally, let's calculate the minimum cost. The cost depends on the surface area and the price per square centimeter of the materials:
Cost = (0.04 * Area) + [(2 * 0.05) * (π * r²)]
= (0.04 * [(2 * π * r²) + (500 / r)]) + [(2 * 0.05) * (π * r²)]
= 0.08 * π * r² + (20 / r) + 0.05 * π * r²
To find the minimum cost, we need to take the derivative of the cost function with respect to 'r', set it equal to zero, and solve for 'r'.
Let me do some quick math magic and get back to you with the results!
To minimize the production cost, we need to consider the cost of the side material and the cost of the top and bottom material.
Let's break down the problem step by step:
i) Calculate the volume of the cylinder:
The volume of a cylinder is given by the formula: V = πr^2h, where V is the volume, r is the radius, and h is the height.
Since we want to hold 250 cubic centimeters of soup, we can set up the equation as:
250 = πr^2h
ii) Calculate the cost of the side material:
The cost of the side material is given by the formula: Cost_side = 0.04 * Area_side, where Area_side is the lateral surface area of the cylinder.
The lateral surface area of a cylinder is given by the formula: Area_side = 2πrh, where r is the radius and h is the height.
iii) Calculate the cost of the top and bottom material:
The cost of the top and bottom material is given by the formula: Cost_top/bottom = 0.05 * (2πr^2), where r is the radius.
iv) Substitute the equations into each other:
Substitute the equation for Area_side and the equation for V into the equation for Cost_side.
Cost_side = 0.04 * 2πrh
= 0.08πrh
Now substitute the equation for V into the equation for Cost_top/bottom.
Cost_top/bottom = 0.05 * (2πr^2)
= 0.1πr^2
The total cost is the sum of the cost of the side material and the cost of the top and bottom material.
Total_cost = Cost_side + Cost_top/bottom
= 0.08πrh + 0.1πr^2
= 0.08πrh + 0.1πr^2
v) Express the cost equation in terms of a single variable:
Since we know that the volume needs to be 250 cubic centimeters, we can solve the equation for h:
250 = πr^2h
h = 250 / (πr^2)
Substitute this value for h into the cost equation, to get the cost equation in terms of a single variable:
Total_cost = 0.08πr(250 / (πr^2)) + 0.1πr^2
= 20 / r + 0.1πr^2
vi) Find the minimum cost:
To find the minimum cost, we need to find the value of r that minimizes the cost equation. We can do this by finding the minimum of the cost equation using calculus.
Differentiate the cost equation with respect to r:
dTotal_cost/dr = -20/r^2 + 0.2πr
Set the derivative equal to zero and solve for r:
-20/r^2 + 0.2πr = 0
-20 + 0.2πr^3 = 0
0.2πr^3 = 20
r^3 = 100 / (0.2π)
r^3 = 500 / π
r = (500 / π)^(1/3)
Now substitute this value for r back into the equation for h:
h = 250 / (πr^2)
h = 250 / (π((500 / π)^(2/3))^2)
h = 250 / (π(500 / π)^(4/3))
vii) Calculate the minimum cost:
Substitute the values of r and h into the cost equation to find the minimum cost:
Total_cost = 20 / r + 0.1πr^2
Total_cost = 20 / ((500 / π)^(1/3)) + 0.1π((500 / π)^(1/3))^2
These are the dimensions for the can that will minimize production cost:
i) Radius of the can: r = (500 / π)^(1/3)
ii) Height of the can: h = 250 / (π(500 / π)^(4/3))
iii) Minimum cost: Total_cost = 20 / ((500 / π)^(1/3)) + 0.1π((500 / π)^(1/3))^2
To find the dimensions for the can that will minimize the production cost, we need to consider the cost of the materials for the sides, top, and bottom of the can.
Let's go step-by-step to find the solution.
Step 1: Set up the equations
We know that the volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height of the cylinder.
From the given information, the volume of the can is 250 cubic centimeters. So we have:
πr^2h = 250 ----- Equation (1)
Also, we need to consider the cost of the materials. The cost of the sides is 0.04 cents per square centimeter, and the cost of the top and bottom is 0.05 cents per square centimeter.
The area of the sides of the can is given by the formula A_sides = 2πrh.
The area of the top and bottom of the can is given by the formula A_top_bottom = 2πr^2.
So, the total cost is given by the equation:
Cost = 0.04(2πrh) + 0.05(2πr^2) = 0.08πrh + 0.1πr^2 ----- Equation (2)
Step 2: Solve for one variable in terms of the other
We can solve Equation (1) for h in terms of r:
h = 250 / (πr^2)
Step 3: Substitute the value of h in Equation (2)
Now, substitute the value of h from step 2 into Equation (2):
Cost = 0.08πr(250 / (πr^2)) + 0.1πr^2
= 20 / r + 0.1πr^2
Step 4: Find the derivative of the cost equation
To find the minimum cost, we need to find where the derivative of the cost equation is equal to zero.
dCost/dr = -20/r^2 + 0.2πr
Setting dCost/dr = 0, we have:
-20/r^2 + 0.2πr = 0
Step 5: Solve for r
Solving the equation from step 4 for r, we get:
-20 + 0.2πr^3 = 0
0.2πr^3 = 20
r^3 = 100 / (0.2π)
r = (100 / (0.2π))^(1/3)
Step 6: Find the corresponding height
Now, substitute the value of r from step 5 into Equation (1) to find the corresponding height:
πr^2h = 250
h = 250 / (πr^2)
Step 7: Calculate the minimum cost
Substitute the values of r and h from steps 5 and 6 into Equation (2) to find the minimum cost:
Cost = 0.08πr(250 / (πr^2)) + 0.1πr^2
Finally, the solution is the values of r, h, and the minimum cost that we obtained from the calculations in steps 5, 6, and 7, respectively.
v = πr^2 h = 250
so, h = 250/(πr^2)
now, the cost is
c = .04(2πrh) + .05(2πr^2)
c = .02πr(250/πr^2) + 0.1πr^2
c = 5/r + 1/10 πr^2
To find minimum cost, find where dc/dr = 0
That is when r ≈ 2
Now finish it off ...