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# Design Problems in Three Dimensions

## Optimization of surface area and volume.

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Practice Design Problems in Three Dimensions
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Design Problems in Three Dimensions

An ice cream cone is made from a thin wafer cookie that is rolled to make the cone shape. The radius of the ice cream cone is 2.5 centimeters and the slant height is 11.5 centimeters. How many square centimeters of wafer cookie are needed to make each ice cream cone?

#### Guidance

When modeling in three dimensions, you are often interested in maximizing or minimizing some aspect of the situation. Perhaps you want to minimize materials used in order to minimize cost. Perhaps you want to maximize the volume in order to maximize profit.

When working with these types of problems, first create your model. Then, test different possibilities or create an equation that shows the relationship between the variable you are looking to maximize or minimize and another variable in the problem.

To find the maximum or minimum value when you know the equation, you can graph the equation and look for where the graph is at a maximum or a minimum.

Example A

A very thin glass container is being designed to hold a candle (it will cover the candle on all sides except the top). The volume of the candle will be 60 cubic inches. If the candle is in the shape of a cylinder, what's an equation that models the amount of glass needed given the radius of the candle?

Solution: Because the glass is very thin, the amount of glass needed is essentially the surface area of the side and bottom of the cylinder.

• The volume of the cylinder is 60 cubic inches, so \begin{align*}\pi r^2 h=60\end{align*}. This means that \begin{align*}h=\frac{60}{\pi r^2}\end{align*}.
• The bottom of the cylinder has a surface area of \begin{align*}\pi r^2\end{align*}.
• The lateral face of the cylinder is a rectangle whose dimensions are the height of the cylinder and the circumference of the base. The surface area of the lateral face is \begin{align*}2 \pi r \left(\frac{60}{\pi r^2}\right)=\frac{120}{r}\end{align*}.

The overall surface area is therefore \begin{align*}SA=\pi r^2+\frac{120}{r}\end{align*}. This is the approximate amount of glass needed for a given radius.

Example B

Using the information from Example A, what should the approximate radius and height of the candle be so that the least amount of glass is used for the container?

Solution: Use a calculator or graphing program to make a graph of the function \begin{align*}SA=\pi r^2+\frac{120}{r}\end{align*}. Look for where there is a minimum in the graph.

The minimum looks to be a radius of approximately 2.6 inches, producing a surface area of approximately \begin{align*}68 \ in^2\end{align*}. When the radius is 2.6 inches, the height will be \begin{align*}h=\frac{60}{\pi r^2}=\frac{60}{\pi (2.6)^2} \approx 2.8 \ inches\end{align*}.

Example C

A new aquarium is designing a central tank. It will be either a cylinder or a square based prism that is in the center of the building, visible from all floors. If the volume of the tank must be 12,000 cubic feet and the height must be 30 feet, should the tank be a cylinder or a square based prism to minimize the glass needed to build the tank? Note that the top of the tank will be open and the bottom will be concrete.

Solution: The glass needed has to do with the surface area of the tank, not including the top or the bottom (which will not be glass).

For the cylinder, the surface area of the lateral face is \begin{align*}2 \pi r h\end{align*}. Since the height must be 30 feet, the surface area is \begin{align*}2 \pi r (30)=60 \pi r\end{align*}. To find the radius, use the fact that the volume must be 12000 cubic feet.

The surface area of the cylinder is \begin{align*}SA_{Cylinder}=60 \pi r=60 \pi (11.28) \approx 2126.23 \ ft^2\end{align*}

For the square based prism, the surface area is the sum of the rectangular faces. If the side of the square base is \begin{align*}s\end{align*}, the surface area is \begin{align*}4(s)(h)\end{align*}. Since the height must be 30 feet, the surface area is \begin{align*}120s\end{align*}. To find the side length, use the fact that the volume must be 12000 cubic feet.

The surface area of the prism is \begin{align*}SA_{Prism}=120s=120 (20)=2400 \ ft^2\end{align*}.

The tank should be a cylinder in order to minimize the amount of glass needed.

Concept Problem Revisited

An ice cream cone is made from a thin wafer cookie that is rolled to make the cone shape. The radius of the ice cream cone is 2.5 centimeters and the slant height is 11.5 centimeters. How many square centimeters of wafer cookie are needed to make each ice cream cone?

When unwrapped, the cone is a sector of a circle. The arc length is the circumference of the base of the cone \begin{align*}(2 \pi r=2 \pi (2.5)=5 \pi \ cm)\end{align*} and the radius is the slant height (11.5 cm).

To find what fraction of the circle this is, note that the circumference of the full circle would have been \begin{align*}2 \pi (11.5)=23 \pi\end{align*}. Therefore, this is \begin{align*}\frac{5 \pi}{23 \pi}=\frac{5}{23}\end{align*} of the circle. To find the area needed for the ice cream cone, find the area of the sector.

#### Vocabulary

Modeling in three dimensions is choosing a solid to represent a real life object and analyzing the solid to help answer questions about the real life object.

Optimization is maximizing or minimizing a particular value.

#### Guided Practice

Use the information from the concept problem for these questions.

1. To save money, you decide you could either make the cones skinnier or shorter. Which change will cause the biggest change in the surface area: shortening the radius to 2 centimeters or shortening the slant height to 11 centimeters?

2. Waffle cones have a radius of 4 centimeters and a slant height of 16 centimeters. How much bigger is the surface area of the waffle cone compared to the surface area of the regular ice cream cone?

3. Come up with a function that outputs the area of wafer cookie needed per cone given the radius and slant height of the cone.

1. If the radius is 2 cm, the new surface area will be \begin{align*}72.26 \ cm^2\end{align*}. If the slant height is 11 cm, the new surface area will be \begin{align*}86.39 \ cm^2\end{align*}. The biggest change in surface area is caused by shortening the radius.

2. The surface area of the waffle cone is \begin{align*}201.06 \ cm^2\end{align*}. This surface area is 110.74 square centimeters bigger than a regular ice cream cone.

3. Let \begin{align*}l=slant \ height\end{align*} and \begin{align*}r=radius\end{align*}. \begin{align*}A=\pi l^2 \cdot \frac{2 \pi r}{2 \pi l}=\pi r l\end{align*}.

#### Practice

1. Come up with at least 5 rectangles with a perimeter of 24 inches. Which rectangle has the biggest area?

2. If you did not consider a square in #1, compare the area of a square with perimeter 24 inches to the other rectangles that you came up with. What do you notice?

3. A pentagon has a given perimeter. Make a conjecture about what type of pentagon with this perimeter will have the biggest area.

4. A rectangular prism has a given surface area. Make a conjecture about what type of rectangular prism will have the biggest volume.

A new peanut butter company wants to stand out compared with the competition, so decides to design a container that is a truncated pyramid. The top of the container is a square that is 4 inches by 4 inches. The bottom of the container is a square that is 2 inches by 2 inches. The container is 4 inches tall.

5. If the pyramid hadn't been truncated, how tall would it have been?

6. Find the volume of the truncated pyramid.

7. Find the surface area of the truncated pyramid.

8. One cubic inch holds 0.45 ounces of peanut butter. How much peanut butter can this container hold?

9. A typical peanut butter jar is a cylinder with a height of 4 inches and a diameter of 3.9 inches. Compare and contrast this type of jar with the new truncated pyramid container.

An open faced box is being made from a square piece of paper that measures 10 inches by 10 inches. The box will be made by cutting small congruent \begin{align*}x\end{align*} by \begin{align*}x\end{align*} sized squares out of each corner.

10. What's an equation that relates \begin{align*}x\end{align*} to the volume of the box?

11. Graph the equation from #10 and explain what it shows.

12. What size squares should you remove from each corner to maximize the volume?

13. Why does it not make sense to try to minimize the volume of the box?

14. The length of the side of the original piece of square paper is \begin{align*}s\end{align*}. Come up with an equation that relates \begin{align*}s\end{align*}, \begin{align*}x\end{align*}, and the volume of the box.

15. Use your equation to verify that a 15 inch by 15 inch piece of paper with 2 inch by 2 inch square corners removed will produce a box with a volume of \begin{align*}242 \ in^3\end{align*}.