Over this semester we have been learning many concepts that all tie into the unit problem. “How soon after they planted the orchard would the center of the lot become a true “orchard hideout”? We have used the Pythagorean Theorem and coordinate geometry, circles, and the square-cube law, and all throughout those two concepts we have been working on giving as much proof as possible for our answers.

The Pythagorean Theorem and Coordinate Geometry

The Pythagorean Theorem is something that plays into very basic math when it comes to right triangles. A right triangle is described as a triangle in which one of the angles is a right angle. The two shorter sides are called the legs and the longest side of a right triangle is called the hypotenuse. The Pythagorean theorem states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides. This is where a^2+b^2=c^2 comes in. The legs added together equal the length of the hypotenuse. The next concept we used that tied into the Pythagorean theorem was coordinate geometry. Coordinate geometry is the study of algebraic equations on graphs. This plays into the Orchard Hideout because we used graphs to plot our points. We also used this concept with the Pythagorean theorem by trying to find things such as the distance between certain points. For example, in one of the warmups, the idea was that for the orchard of radius 3 the best line of sight passed through the lower gap. The perpendicular distance from the tree at C (1,0) to that line was approximately 0.32 units. The question, however, was asking about the other gap that was shaded on the top quadrant of the gap that was bounded by trees A and B. The idea similar to the first prompt was to be able to find the distance from tree A which was at (1,1) to the optimal line of sight through the gap. We determined that the optimal line of sight went through point (3,2) by drawing a right triangle and using those two points to determine the distance from tree A to our optimal line of sight at (3,2). To solve this problem we also used the distance formula: D=(x2-x1)2+ (y2-y1)2.After using the Pythagorean Theorem, Coordinate Geometry, and the distance formula we determined that the distance between our original point and our optimal line of sight is 3.

Circles and the Square-Cube Law

The square-cube law appears everywhere whether it be the physical properties of stars and planets or the spread of infectious diseases. The most important aspect of the square-cube law, however, is how strength scales with size. The Square-cube law states that area scales with the square of length while volume scales with the cube of length. In “The Orchard Hideout: The Square-Cube Law and Unpacking the article”, we tested this theory. We had two spheres, one with a radius of 2 cm, and one with a radius of 10 cm. The formula for surface area is 4r2. When you plug in the numbers you get 50.26 for the 2 cm sphere and for the 10 cm sphere the surface area equaled 1,256.6. The next problem was to be able to find the volume of each sphere. The formula for finding the volume of a sphere is 43r3 . When we plugged in the numbers for the 2 cm sphere the result was 33.5 cm. When we did the same for the 10 cm sphere the result was 4,188.7 cm. This worksheet also gave us the formulas for a circle. D= 2r, C= 2r, and A=r2. This concept applies to circles because whatever size circle you have you’re going to need to know how to find its surface area and volume and this concept helps greatly with that.

Proof

Providing proof is one of the most important aspects of mathematics. Sometimes, however, it seems very difficult to provide proof. If you show someone a math problem and the solution to that math problem and they ask “Why?’ you might feel stuck and not know what to say or do. However, one of the most helpful things in mathematics is “Proofs.” A mathematical proof is an inferential argument for a mathematical statement, showing that the stated assumptions logically guarantee the conclusion. A couple of the proofs we went over were “Two ways to derive the formula for the area of a circle”. In the first video we watched it was more of a visual proof. Essentially if you take a circle and draw more circles inside of the original circle and then “stretch” the other circles into a straight line it will create something that looks like a bar graph. This bar graph will also represent a triangle because the lines get smaller and smaller when you continue to draw them inside the original circle. The base of this “bar graph” will represent the radius, the height will represent the circumference and the hypotenuse will represent the area. In the second proof, the man talks about how if we take a rectangle, the area of that rectangle will equal the area of the circle and we can determine this by cutting the circle into sections and taking those sections and making a triangle. The smaller the pieces, the more it resembles a triangle. So if the circle was completely filled in and there were so many lines that they were indistinguishable, that is when the pieces of the circle would most represent a triangle.

These two proofs showed us how to find the area of a circle and be able to prove it at the same time. Proof was very important in solving the unit problem because we needed to be sure to show our work and be able to prove that our solution to the unit problem was correct.

Introduction

“How soon after they planted the orchard would the center of the lot become a true “orchard hideout”? This is the unit problem that we have been leading up to throughout the semester. Using the Pythagorean Theorem and coordinate geometry, the Square-Cube law and circles, and Proofs to be able to solve this problem. The final objective was to be able to solve this problem using all of the concepts that we have learned this semester.

Process and Justification

The first thing that needed to be looked at was all the information they gave us to be able to solve this problem. The problem states that the cross-sectional area of a tree trunk increases by 1.5 square inches per year, currently, the tree trunks each have a circumference of 2.5 inches, the unit distance [for instance, the distance from (0, 0) to (1, 0)] is 10 feet, and, the last line of sight is the line that goes from the origin through the point (25, 12). Another important thing to keep in mind is that they planted their orchard on a circular lot of radius 50 units. When we look at the points on the graph we can see that the line of sight needs to go through point (25, 12) which is the midpoint. Therefore, our last line of sight also needs to go through point (50,1). Another important thing that we know is that the final radius of our tree comes from the distance from a point to a line so we are looking for the perpendicular distance. What we need to do next is create two triangles. A triangle with a right angle against our line of sight and a bigger triangle that goes out to the end of the orchard and follows the line of sight. The big triangle is 50 on the base and one on the vertical end and the small triangle has a hypotenuse of length 1 and we need to be able to find the length of the vertical end. The next step is to solve for the hypotenuse of the big triangle so we follow the Pythagorean theorem a2 + b2=c2. So we take 50 squares plus 1 squared which equals 2,500. We then add 2,500 + 1 which gives us 2,501. After all of the addition, we then need to find the square root of that 2,501 to give us c2. After we square root 2,501 we get 50.009. The next thing we need to do is set up our ratio to find the missing vertical side of the small triangle. If we think about it the missing vertical side of the small angle (lets call it n) divided by one is going to be the same as 1 divided by the hypotenuse. So n is going to equal 12,501 once we do that equation we can determine that n equals about 0.019996 units which is the radius of our hideout.

The next thing we need to do is find the starting and ending area for the trees so that we can figure out how many years it will take to for the orchard to become a true hideout. We know that the trees start at 2.5 inches in circumference and they grow by 1.5 square inches per year so we need to find the radius of those trees. After using the radius formula we can determine that the radius of the trees at the start of their growth equal 0.39 in. Next, we need to find the area so when we plug in the area formula and the corresponding numbers we can determine that the area of the starting trees is equal to 0.497 inches squared. To find the ending area of the trees we need to use the radius which is 0.019996 units. We then take the radius and and multiply it with the idea that there are 10 feet in one unit and then we multiply that with the idea that there are 12 inches in one foot. After doing that all of the units and feet cancel out so we are left with 0.019996 times 10 times 12 inches. After doing that multiplication we get a radius of 2.4 inches. Then we need to find the area again by using the area formula and we get 18.08 in squared. Next er need to take the starting area and the ending area and see how long it takes to get from each point when the trees are increasing by 1.5 inches per year or in other words, we need to find the difference by subtracting 0.497 from 18.08 which gives us 17.59 inches squared. Now we have to divide how much the tree needs to grow by how fast the tree needs to grow. After we do this we get our answer which is 11.73 years. So the orchard will take 11.73 years to become a true hideout.

Say Leslie were to plant two flowers and let’s say Leslie also has a sprinkler that sprays around in a circle. Leslie wants to place the sprinkler where it will equally distribute water in between the two flowers. In order for the flowers to receive the same amount of water equal distance from the two flowers. The easiest solution is to place the sprinkler directly in the center between the two flowers but let’s say she wants to place the sprinkler somewhere else. Is this possible? The answer is yes. As long as the sprinkler is vertical from the two flowers it will always get the same amount of water no matter how far away it is. The reason behind this is because if you were to put this scenario on a graph and draw it out you would see that if the sprinkler is vertical from the flowers is creates a right triangle. If you draw a circle around the two flowers and the sprinkler you will see that the sprinkler is on the perpendicular bisector. So the solution to two flowers is to make sure the sprinkler is always on the perpendicular bisector that way the two flowers are always getting the same amount of water.

Now say Leslie wanted to plant three flowers instead of two and she also wanted to put the same sprinkler in a place where each flower gets the same amount of water. So where would she put it this time? The solution to this question is essentially the same as the last solution. Let’s say you were to draw this out on a piece of paper. You would draw three dots representing the flowers. Now connect those three dots and what do you get? A triangle. Now if you draw a circle around the triangle and put a dot right in the center of the triangle and circle you can see that if you put the sprinkler on the perpendicular bisector it will be in the center of the flowers and it will distribute the water evenly amongst the flowers.

The solution will always be the same no matter how many flowers are. As long as the Sprinkler lands on the perpendicular bisector you can count on the sprinkler reaching all the flowers in even amounts. As long as the sprinkler sprays infinitely there is no limit to the number of flowers that she can plant. However, if the sprinkler only sprayed a certain amount of feet then that would be a completely different scenario and the number of flowers that could be planted would be limited by the length of the sprinkler but as long as the sprinkler length is infinite and the sprinkler lands on the perpendicular bisector there is no limit to the number of flowers that Leslie can plant.

The Orchard Hideout: Another Kind of Bisector

In our work on The Sprinkler Dilemma, we came across the idea of a perpendicular bisector, which is a particular line that splits a line segment into two equal parts. But line segments aren’t the only geometric figures that have bisectors. In Fire! Fire! we are learning about a new kind of bisector.

An angle bisector is a ray that splits an angle into two equal parts. For example, in the diagram at the right, if ∠RST = 50° and angles RSU and UST each equal 25°, then ray SU is the bisector of ∠RST.

Suppose two lines l and m intersect at point A, as shown in the diagram below. In this diagram, B and D are two points on l, and C and E are two points on m. The intersecting lines l and m form four angles at A, namely, ∠BAC, ∠CAD, ∠DAE, and ∠EAB.

- The diagram shows that ∠BAC = 70°. Find the size of each of the other three angles.

DAE= 70 degrees

CAD= 110 degrees

EAB=110 degrees

- In the diagram below, the dashed rays represent the four angle bisectors. That is, ray AP bisects ∠BAC, ray AQ bisects ∠CAD, ray AR bisects ∠DAE, and ray AS bisects ∠EAB.

- Prove that the two angle bisectors, rays AP and AR are part of the same line. That is, show that points P, A, and R are collinear. (Hint: Use the fact that ∠BAC = 70° to show that ∠PAR = 180°.)

- Prove that the angle bisectors AQ and AS are part of the same line.

- Prove that the line containing AP and AR is perpendicular to the line containing AQ and AS.

AP is perpendicular to AR because when they intersect each other they create a 90 degree angle

- Your work in Question 2 was based on the fact that ∠BAC = 70°. Now represent ∠BAC as x and prove parts a, b, and c of Question 2 in general. (Hint: Express other angles in terms of x.)

x2+(180-x) + x2

=x+180-x

=180 degrees

I feel like as a mathematician I grew quite a bit during this unit. I have always struggled with math and there are definitely some things I still need to work on but I do feel like I learned a lot this semester. From expanding my knowledge on the Pythagorean Theorem to learning totally new things such as the square-cube law. One thing I still really need to work on is asking questions when I am confused and need help. I feel as though that’s where I struggled most this semester and my role in not asking questions caused me to suffer in some of the assignments.

The unit improved my understanding of the relationship between algebra and geometry because when we learned new concepts we went through them in a very detailed way and went through every step one by one. Sometimes it was a little overwhelming for me but I got to learn how we can use lines and shapes on a graph to find something like how long it would take the orchard hideout to become a true hideout. I look forward to next semester and I will be sure to ask more questions when I am confused or need help.

Writeup

According to the corner point principle, the optimal production policy is represented by a feasible region. To determine the optimal production policy, we have to find the corner points of our region and evaluate the profit relation. In this unit problem we needed to find the most reasonable costs for Dalton Ranch, The Old Fort, and The Boston mine. The costs for these lands could either be for recreation or development. The costs are shown in the table below.

Parcel

Improvement costs per acre for recreation

Improvement costs per acre for development

Dalton Ranch

$50

$500

The Old Fort

$200

$2,000

The Boston Mine

$100

$1,000

We started off by going through each of the 12 constraints to see if we could eliminate any of them. The constraints are listed below.

- RR + RD = 300
- AR + AD = 100
- MR + MD = 150
- RD + AD + MD ≥ 300
- AR + MR ≤ 200
- AR + RD = 100
- RR ≥ 0
- AR ≥ 0
- MR ≥ 0
- RD ≥ 0
- AD ≥ 0
- MD ≥ 0

After going through each constraint we were able to eliminate three. VII, X, and, XI. After eliminating those three constraints we took the remaining 9 and came up with 10 possible combinations that would result in the least amount of money paid for these three landscapes. The combinations are listed below.

- I, II, III, VI, IV, V
- I, II, III, VI, IV, VIII
- I, II, III, VI, IV, IX
- I, II, III, VI, IV, XII
- I, II, III, VI, V, VIII
- I, II, III, VI, V, IX
- I, II, III, VI, V, XII
- I, II, III, VI, VIII, IX
- I, II, III, VI, VIII, XII
- I, II, III, VI, IX, XII

After finding each combination we found the matrices using multiplication.

After plugging the combinations into the matrix calculator and using inverse operations were able to find three combinations that would work the best. These combinations were numbers 2, 3, and 8. We then took the costs of the lands and multiplied them by the matrices that we got. The totals are listed above in the previous pictures. We were able to determine that combination three resulted in the least amount of money spent.

Reflection

I think that a key insight I gained about myself is that if I really pay attention and really work hard I have an easier time understanding the math that we are learning and I feel less stressed. I thought the content in this unit was confusing at first and there was a lot of information at one time but once I understood how to do it, I felt that it became easier. I also enjoyed multiplying matrices and plugging them into the matrix calculator. It was nice that the calculator did everything for you and all you had to do was plug in the numbers.

I am not proud of my work from the beginning of the unit because I was confused and I didn’t really understand how to do it but I am proud of my work that was from the end of the unit. I feel as though if you were to look at my overall work from the unit you would be able to see that progression. I have learned that I sometimes need things explained to me more than once or in a different way but once I understand it, it sticks. I have developed my habits as a mathematician by asking lots of questions when I am confused. It helps me to feel more confident in the work I am doing.