A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 550 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.02 cents per square centimeter. The top will be made of glued paper, costing 0.07 cents per square centimeter. Find the dimensions for the package that will minimize production cost.
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. Since we are given that the volume of the container is 550 cubic centimeters, we have:
550 = πr^2h
We want to minimize the cost of production, which is the sum of the costs for the sides and bottom (styrofoam) and the cost for the top (paper). The cost for the sides and bottom is given by:
Cost_side+bottom = 0.02(2πrh + πr^2)
The cost for the top is given by:
Cost_top = 0.07(πr^2)
The total cost is therefore:
Cost_total = Cost_side+bottom + Cost_top
Cost_total = 0.02(2πrh + πr^2) + 0.07(πr^2)
We can substitute the expression for the volume into the cost equation to get a cost function in terms of one variable. Since we want to minimize the cost, we will take the derivative of the cost function with respect to the variable r and set it equal to zero to find the value of r that minimizes the cost.
After solving for r, we can then find the value of h using the volume equation.
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anonymous
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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. Since we are given that the volume of the container is 550 cubic centimeters, we have:
550 = πr^2h
We want to minimize the cost of production, which is the sum of the costs for the sides and bottom (styrofoam) and the cost for the top (paper). The cost for the sides and bottom is given by:
Cost_side+bottom = 0.02(2πrh + πr^2)
The cost for the top is given by:
Cost_top = 0.07(πr^2)
The total cost is therefore:
Cost_total = Cost_side+bottom + Cost_top
Cost_total = 0.02(2πrh + πr^2) + 0.07(πr^2)
We can substitute the expression for the volume into the cost equation to get a cost function in terms of one variable. Since we want to minimize the cost, we will take the derivative of the cost function with respect to the variable r and set it equal to zero to find the value of r that minimizes the cost.
After solving for r, we can then find the value of h using the volume equa
I will be happy to help you further with the calculations to find the dimensions that minimize the production cost of the cup-of-soup package. Let's continue from the point where we need to take the derivative of the cost function with respect to the variable r and set it equal to zero to find the optimal value for r.
Starting with the total cost function we derived:
Cost_total = 0.02(2πrh + πr^2) + 0.07(πr^2)
Substitute the volume equation 550 = πr^2h into the cost equation:
Cost_total = 0.02(2πr(550/r) + 550) + 0.07(550)
Cost_total = 0.04π(550) + 0.02πr(550) + 0.07(550)
Now simplify the equation:
Cost_total = 22π + 11πr + 38.5
To minimize the cost, take the derivative of Cost_total with respect to r and set it equal to zero:
d(Cost_total)/dr = 11π = 0
Solving for r:
11π = 0
r = 0
This value does not make sense in this context since the radius cannot be zero. It seems there might have been a mistake in the calculations. Let's review the process and correct any errors to find the correct dimensions that minimize the production cost.