# A microwaveable cup-of-soup package needs to be constructed in the shape of a cylinder to hold 250 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.03 cents per square centimeter. The top will be made of glued paper, costing 0.08 cents per square centimeter. Find the dimensions for the package that will minimize production costs.

To minimize the cost of the package:

Radius =?

Height=?

Minimum cost=?

## To find the dimensions for the package that will minimize production costs, we need to minimize the cost function.

Let's denote the radius of the package as "r" and the height as "h".

The volume of the cylinder is given as 250 cubic centimeters, so we have the equation:

V = πr^2h = 250

To minimize the production costs, we need to minimize the cost function.

The cost of the sides and bottom of the container is given by:

Cost of styrofoam = surface area of sides and bottom * cost per square centimeter

= 2πrh + πr^2 * cost per square centimeter

= 2πrh + πr^2 * 0.03

The cost of the top of the container is given by:

Cost of paper = surface area of top * cost per square centimeter

= πr^2 * cost per square centimeter

= πr^2 * 0.08

The total cost function is the sum of the cost of the sides/bottom and the cost of the top:

Cost = 2πrh + πr^2 * 0.03 + πr^2 * 0.08

= 2πrh + πr^2 * 0.03 + πr^2 * 0.08

= 2πrh + πr^2 * (0.03 + 0.08)

= 2πrh + πr^2 * 0.11

To minimize the cost, we can take the derivative of the cost function with respect to "r" and "h", and set it equal to 0.

∂Cost/∂r = 0

∂Cost/∂h = 0

Taking the derivative with respect to "r":

∂Cost/∂r = 2πh + 2πr * 0.11 = 0

Simplifying:

2πh + 0.22πr = 0

h + 0.11r = 0

Taking the derivative with respect to "h":

∂Cost/∂h = 2πr = 0

Simplifying:

2πr = 0

Since 2πr = 0 doesn't give us any meaningful information, we can ignore it.

Now, we can solve the system of equations:

h + 0.11r = 0 (1)

2πrh + πr^2 * 0.11 = 0 (2)

V = πr^2h = 250 (3)

Solving equation (1) for h:

h = -0.11r (4)

Substituting equation (4) into equation (3):

V = πr^2 * (-0.11r) = 250

Simplifying:

-0.11πr^3 = 250

Solving for r:

r^3 = -250/-0.11π

r^3 = 2272.727

Taking the cube root:

r ≈ 13.39 cm

Substituting the value of r back into equation (4) to find h:

h = -0.11 * 13.39 ≈ -1.47 cm

Since the height cannot be negative, we can take the absolute value:

h ≈ 1.47 cm

The dimensions for the package that will minimize production costs are:

Radius ≈ 13.39 cm

Height ≈ 1.47 cm

To find the minimum cost, substitute the values of r and h back into the cost function:

Cost = 2πrh + πr^2 * 0.11

Cost ≈ (2π * 13.39 * 1.47) + (π * 13.39^2 * 0.11)

Cost ≈ 102.14 + 623.67

Minimum cost ≈ 725.81 cents

Therefore, the minimum cost for the package is approximately 725.81 cents.

## To minimize the cost of the package, we need to find the dimensions (radius and height) that will minimize the total cost of both the styrofoam and the paper.

Let's assume the radius of the cylinder is "r" cm, and the height is "h" cm.

The volume of the cylinder can be calculated using the formula:

Volume of cylinder = πr^2h

Given that the volume of the cup-of-soup package is 250 cubic centimeters, we have:

250 = πr^2h

Next, we need to calculate the areas of the different parts of the container to find the cost.

The cost of the styrofoam sides and bottom is given by:

Styrofoam cost = (area of sides + area of bottom) × 0.03 cents

The area of the sides of the cylinder can be calculated using the formula:

Area of sides = 2πrh

The area of the bottom of the cylinder can be calculated using the formula:

Area of bottom = πr^2

The cost of the paper top is given by:

Paper cost = area of top × 0.08 cents

The area of the top of the cylinder can be calculated using the formula:

Area of top = πr^2

Finally, the total cost of production is the sum of the styrofoam cost and the paper cost:

Total cost = (Styrofoam cost + Paper cost)

To minimize the total cost, we need to find the values of "r" and "h" that satisfy the volume equation and calculate the minimum cost.

To solve this problem, we can use calculus to find the minimum cost.

1. Substitute the volume equation into the cost equation to obtain the total cost in terms of one variable, either "r" or "h".

2. Take the derivative of the total cost equation with respect to the chosen variable.

3. Set the derivative equal to zero and solve for the variable.

4. Substitute the value of the variable back into the volume equation to find the corresponding value of the other variable.

5. Calculate the minimum cost by substituting the values of "r" and "h" into the total cost equation.

Alternatively, you can use numerical methods or graphing software to plot the cost function and find the minimum point.

Remember to consider constraints such as practical limitations on the dimensions of the cup-of-soup package when interpreting the results.

Please note that the calculations involved in finding the precise values for "r," "h," and the minimum cost require further steps and computations that go beyond the scope of a simple explanation.

## I will skip the units for easier typing

let the radius be r, let the height be h

Need: V = 250

πr^2 h = 250

h = 250/(πr^2)

cost = .03(πr^2) + .03(2πrh) + .08(πr^2)

= .11π r^2 + .06πr(250/(πr^2)

= .11πr^2 + 27.5/r

d(cost)/dr = .22πr - 27.5/r^2 = 0 for a min of cost

.22πr = 27.5/r^2

r^3 = 27.5/(.22π) = 39.7887..

r = 3.4139.. cm

h = 6.8278.. cm , (notice h = 2r)

sub those into the cost equation to find the min cost

Your answer will be in cents.