# Eureka Math Algebra 2 Module 1 Lesson 16 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 16 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 16 Mathematical Modeling Exercise Answer Key

You will be assigned to a group, which will create a box from a piece of construction paper. Each group will record its box’s measurements and use said measurement values to calculate and record the volume of its box. Each group will contribute to the following class table on the board.

Using the given construction paper, cut out congruent squares from each corner, and fold the sides in order to create an open-topped box as shown on the figure below.

Exercise 1.
Measure the length, width, and height of the box to the nearest tenth of a centimeter.
Answer:
Answers will vary.
Length: L = 35.7 cm
Width: W = 20.5cm
Height: H = 5.0 cm

Exercise 2.
Calculate the volume.
Answer:
Answers will vary. Sample answer:
Volume: V = L ∙ W ∙ H = 3,659.25cm3

Exercise 3.
Have a group member record the values on the table on the board.
Answer:
Discuss the results, and compare them with the conjectures made before cutting the paper.
→ Who was able to enclose the most volume?
→ Why would our goal be to enclose the most volume?
We are optimizing our resources by enclosing more volume than the other groups using the same-size paper.
Have students continue with the exercise.

Exercise 4.
Create a scatterpiot of volume height using technology.
Answer:

Exercise 5.
What type of polynomial function could we use to model the data?
Answer:
A cubic or quadratic polynomial; we cannot tell from just this portion of the graph.

Exercise 6.
Use the regression feature to find a function to model the data. Does a quadratic or a cubic regression provide a better fit to the data?
Answer:
Answers will vary based on the accuracy of the measurements, but the cubic regression should be a better fit.
Sample answer: V(x) = 4x3 – 152. 4x2 + 1,398. 8x

Exercise 7.
Find the maximum volume of the box.
Answer:
The maximum volume is 3,770.4 cm3.

Exercise 8.
What size square should be cut from each corner in order to maximize the volume?
Answer:
A 6 cm × 6 cm square should be cut from each corner.

Exercise 9.
What are the dimensions of the box of maximum volume?
Answer:
The dimension are 33.7 cm × 18.5 cm × 6 cm

### Eureka Math Algebra 2 Module 1 Lesson 16 Problem Set Answer Key

Question 1.
For a fundraiser, members of the math club decide to make and sell “Pythagoras may have been Fermat’s first problem but not his last!” t-shirts. They are trying to decide how many t-shirts to make and sell at a fixed price. They surveyed the level of interest of students around school and made a scatterplot of the number of t-shirts sold (x) versus profit shown below.

a. Identify the y-intercept. Interpret Its meaning within the context of this problem.
Answer:
The y-intercept Is approximately -125. The -125 represents the money that they must spend on supplies in order to start making t-shirts. That is, they will lose $125 if they sell 0 t-shirts. b. If we model this data with a function, what point on the graph of that function represents the number of t-shirts they need to sell in order to break even? Why? Answer: The break-even point Is the first x-intercept of the graph of the function because at this point profit changes from negative to positive. When profit is 0, the club is breaking even. c. What is the smallest number of t-shirts they can sell and still make a profit? Answer: Approximately 12 or 13 t-shirts d. How many t-shirts should they sell in order to maximize the profit? Answer: Approximately 35 t-shirts e. What is the maximum profit? Answer: Approximately$280

f. What factors would affect the profit?
Answer:
The price of the t-shirts, the cost of supplies, the number of people who are willing to purchase a t-shirt

g. What would cause the profit to start decreasing?
Answer:
Making more t-shirts than can be sold

Question 2.
The following graph shows the temperature in Aspen, Colorado during a 48-hour period beginning at midnight on Thursday, January 21, 2014. (Source: National Weather Service)

a. We can model the data shown with a polynomial function. What degree polynomial would be a reasonable choice?
Answer:
Since the graph has 4 turning points (2 relative minima, 2 relative maxima), a degree 5 polynomial could be used. Students could also argue that the final point is another minimum point and that a degree 6 polynomial could be used.

b. Let T be the function that represents the temperature, in degrees Fahrenheit, as a function of time t, in hours. If we let t = 0 correspond to midnight on Thursday, interpret the meaning of T(5). What is T(5)?
Answer:
The value T(5) represents the temperature at 5 a.m. on Thursday. From the graph, T(5) = 13.

c. What are the relative maximum values? Interpret their meanings.
Answer:
The relative maximum values are approximately T(13) = 28 and T(37) = 34. These points represent the high temperature on Thursday and Friday and the times at which they occurred. The high on Thursday occurred at 1: 00 (when t = 13) and was 2 8°F. The high on Friday occurred at 1: 00 (when t = 37) and was 34°F.

### Eureka Math Algebra 2 Module 1 Lesson 16 Exit Ticket Answer Key

Jeannie wishes to construct a cylinder closed at both ends. The figure below shows the graph of a cubic polynomial function used to model the volume of the cylinder as a function of the radius if the cylinder is constructed using 150π cm2 of material. Use the graph to answer the questions below. Estimate values to the nearest half unit on the horizontal axis and the nearest 50 units on the vertical axis.

Question 1.
What is the domain of the volume function? Explain.
Answer:
The domain is approximately 0 ≤ r ≤ 8.5 because a negative radius does not make sense, and a radius larger than 8.5 gives a negative volume, which also does not make sense.

Question 2.
What is the most volume that Jeannie’s cylinder can enclose?
Answer:
Approximately 800 cm3

Question 3.
What radius yields the maximum volume?
Answer:
Approximately 5 cm

Question 4.
The volume of a cylinder is given by the formula V = πr2h. Calculate the height of the cylinder that maximizes the volume.
Answer:
Approximately 10.2 cm

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