Optimal+Values



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Optimal Values include perimeter, area, volume and surface area. Perimeter and Area are used in 2D shapes, such as the rectangle and the triangle. Volume and Surface Area are used in 3D shapes, such as the rectangular prism and the cylinder. In Optimal Values, there are definitions, explanations and examples for all four topics.


 * **Area**: the number of square units in a region; measured in units squared
 * **Perimeter**: the distance around a polygon; measured in units
 * **Surface area**: The sum of the area on the outside of a three-dimensional shape; measured in units squared (u²)
 * **Volume**: the amount of space that an object occupies; measured in units cubed (u³)
 * ** π:** Pi is the ratio of the circumference of a circle to its diameter (If the diameter of the circle is 1, the circumference is pi). It is approximately 3.14159. We will use it in this section to calculate the optimal values of circles.
 * **Diameter:** A straight line segment that passes through the center of a circle, and ends on the edge of the circle. It is the max distance from one end of a circle to another.
 * **Radius:** A straight line segment that starts at the center of a circle and ends on an edge. It is always half of the diameter.
 * **Pythagorean theorem:** basically stated as the sum of the areas of two small squares = the area of the large one. It is used for calculating the unknown side in a right-angle triangle, the formula being [[image:3ae71ab3eb71d3d182a3b9e437fba6ee.png width="80" height="15"]]
 * These are the formulas that will be covered in Optimal Values: **



**2D Shapes that will be covered:** 
 * //Rectangle//
 * //Triangle//
 * //Trapezoid//
 * //Circle//
 * 2D shapes use perimeter and area.**



====**Perimeter is the distance around a two-dimensional shape. To find the perimeter of an object, you add the measurements of all the sides together. Perimeter is measured in units. The perimeter formula will differ between shapes however, a formula that stays constant is //P//=sum of all sides. For instance, a rectangle has 2 pairs of equal sides with equal lengths. So, the formula would be //P//= side 1(2) + side 2(2) or //P//=sum of all sides. However, an equilateral triangle has 3 equal sides with equal lengths. So, the formula would be //P//= sides(3), or //P//=sum of all sides. For a square the formula would be P= L(4).**====

__**Example 1:**__

Calculate the perimeter of the rectangle. The perimeter of the rectangle is 18 cm.

__**Example 2: Perimeter of a square.**__ Because all four sides of a square are the same length, you can multiply the length of one side by four. In this case, that would be 2 x 4. 2 cm

P= //s//+//s//+//s//+//s// P= 4//s P= 4(2) P= 8 // the perimeter is 8 cm. Area is the space inside a two-dimensional shape. The formula for area varies between shapes. Area is measured in squared units. You can calculate area in many different ways. The area of a rectangle is calculated by using the formula //A//=//LxW//.

__**Example 1:**__

Calculate the area of the rectangle.

 The area of the rectangle is 18 cm ² .

__**Example 2:**__

Calculate the area of the square. <span style="font-family: Verdana,Geneva,sans-serif;"> The area of the square is 16 cm <span style="color: rgb(23, 74, 136); font-family: Verdana,Geneva,sans-serif;">².


 * __<span style="color: rgb(255, 130, 0); font-size: 120%;"> CHECK YOUR UNDERSTANDING (A):__**

<span style="color: rgb(128, 0, 128);"> Determine the area and perimeter of the shape below. Answers are located above "citations".

__**Minimum Perimeter and Maximum Area**__

When given a certain area, we can find the least amount of perimeter required to create that area. When given a certain perimeter, we can find the maximum area that perimeter can create.

__**Example 1:**__

Find the minimum perimeter of a rectangle with an area of 48cm².



The formula to find the minimum perimeter of a rectangle is: //P=// (√//A//)(4).

So, the minimum perimeter of a rectangle with an area of 48cm² is:

//P=// (√//A//)(4) //P=// (√//48)//(4) //P= (6.93)(4) P= 27.72cm//

<span style="font-family: Verdana,Geneva,sans-serif;">The minimum perimeter of a rectangle with an area of 48cm² is 27.72cm.


 * The shape that creates the minimum perimeter of a rectangle will always be a square.**

__**Example 2:**__

Find the maximum area of a rectangle with a perimeter of 24cm.



The formula to find the maximum area of a rectangle is: //A= (P/4)²//.

So, the maximum area of a rectangle with a perimeter of 24cm is:

//A= (P/4)² A= (24/4)² A= 6² A= 36cm².// <span style="font-family: Verdana,Geneva,sans-serif;"> The maximum area of a rectangle with a perimeter of 24cm is 36cm².


 * The shape that creates the maximum area of a rectangle will always be a square.**

<span style="font-family: Verdana,Geneva,sans-serif;">**Volume and Surface Area are used by 3D shapes.**

Volume is the amount of space occupied by a 3-D shape. Volume is measured in cubic units. Depending on the object, the volume of a shape can differ, along with the formula. For example, the formula for finding the volume of a cylinder is **//V=// π**<span style="font-size: 110%; font-family: Verdana,Geneva,sans-serif;">**r**, a**²** **h.** The fomula for volume of a rectangular prism is **//V//=** //**lwh**.//

To calculate the volume of a <span style="color: rgb(255, 0, 174); font-family: Tahoma,Geneva,sans-serif;">Triangular Prism you use the formula base times length times height, then divide by two: <span style="font-size: 300%; color: rgb(95, 0, 255); font-family: Verdana,Geneva,sans-serif;"> <span style="font-size: 900%; color: rgb(95, 0, 255); font-family: Verdana,Geneva,sans-serif;"> <span style="font-size: 300%; color: rgb(95, 0, 255); font-family: Verdana,Geneva,sans-serif;"> Therefore, the volume is 3,000ft. To find the volume for a cylinder, you use the formula:** <span style="font-size: 300%; color: rgb(23, 74, 136); font-family: Verdana,Geneva,sans-serif;"> <span style="font-size: 300%; color: rgb(95, 0, 255); font-family: Verdana,Geneva,sans-serif;">
 * __Example 1:__**
 * <span style="color: rgb(0, 0, 0);">__Example 2:__
 * <span style="color: rgb(0, 2, 255);">V= πrh **

V= **<span style="color: rgb(206, 166, 166); font-family: Arial,Helvetica,sans-serif;">**<span style="color: rgb(0, 2, 255);">603.19cm³ ** <span style="font-size: 300%; color: rgb(95, 0, 255); font-family: Verdana,Geneva,sans-serif;"> **__Example 3:__ Volume of a Rectangular Prism: V=2(wh+lw+lh) V=2(6)(5)+(4)(6)+(4)(5) V=2(30)+(24)+(20) V=2(74) V=148cm³ Therefore, the volume is 148cm³**
 * <span style="color: rgb(95, 0, 255);">V= <span style="color: rgb(23, 74, 136);">π(4)²(12)
 * <span style="color: rgb(0, 2, 255);">Therefore, the volume is 603.19cm <span style="color: rgb(206, 166, 166);">³ **



<span style="font-family: Verdana,Geneva,sans-serif;">Surface area is the number of square units needed to cover the whole surface of a three-dimensional object. You can calculate the surface area of an object by finding the areas of all the faces on the shape and adding the areas together.

//**__Surface Area of a <span style="color: rgb(214, 46, 46);">Cylinder = <span style="color: rgb(46, 121, 37);">2πr²+2πrh Surface Area of a <span style="color: rgb(237, 44, 44);">Rectangular prism = <span style="color: rgb(59, 148, 46);">2(wh+lw+lh) Surface Area of a <span style="color: rgb(231, 54, 54);">Triangular Prism =<span style="color: rgb(55, 124, 45);"> ah+bh+ch+bl

Example: Cylinder__**// H=15yd

<span style="color: rgb(15, 69, 199); font-family: Verdana,Geneva,sans-serif;">SA=2πr²+2πrh SA= 2π(9)²+2π(9)(15) SA=508.94+848.23 SA=1357.17yd² <span style="font-family: Verdana,Geneva,sans-serif;"> <span style="color: rgb(233, 78, 224); font-family: Verdana,Geneva,sans-serif;">Therefore the surface area of the cylinder is 1357.17yd²

__**Example # 1

**__ <span style="font-family: Verdana,Geneva,sans-serif; color: rgb(95, 0, 255);"> SA= 2bs + b² SA= 2(4)(3.18) + 4² SA= 25.44 + 16 SA= 41.44 m² <span style="font-family: Verdana,Geneva,sans-serif; color: rgb(0, 255, 255);">Therefore the surface area of the square-based pyramid is 41.44m²


 * __Example # 2__**

<span style="font-family: Verdana,Geneva,sans-serif;"> <span style="font-family: Verdana,Geneva,sans-serif; color: rgb(0, 0, 255);">SA= 4πr² SA= 4π(14)² SA= 4π(196) SA= 2463.01 <span style="font-family: Verdana,Geneva,sans-serif; color: rgb(128, 0, 0);">Therefore the surface area of the sphere is 2463.01 cm² <span style="font-family: Verdana,Geneva,sans-serif; color: rgb(0, 0, 255);">

<span style="color: rgb(255, 0, 174);"> When given a certain area of a square, we can find the minimum perimeter that will give us that area. When given its perimeter, we can find the maximum area for that perimeter.

__**<span style="color: rgb(255, 130, 0); font-size: 120%;"> CHECK YOUR UNDERSTANDING (B):**__

<span style="color: rgb(0, 128, 128);">Determine the volume and surface area of the shape below. Answers are located above "citations". **REMEMBER**: //r//= //d///2



__**Pythagorean Theorem**__

The Pythagorean Theorem is important because we can use it to find any side of a right triangle, as long as we are given the two other sides. This can be useful to solve some unknowns. Take a look at the diagram:



Pythagorean Theorem: In any right triangle, the area of the hypotenuse (side opposite of the right angle) squared is equal to the area of the other two sides squared.

The Pythagorean Theorem can be written as:

A² + B² = C²
//C// is the hypotenuse, //A and B// are the two legs (sides that meet at the right angle) of the triangle.

Using this, we can sometimes solve an unknown value in a triangle, cone, square-based pyramid or triangular prism, which might allow us to find the area/perimeter/surface area/volume.

__**Example 1:**__

Find the length of the missing side.



Since we are trying to find the hypotenuse, the formula to be used is: A² + B² = C²

A² + B² = C² 12² + 12² = C² 144+144 = C² 288 = C² √288 = C C = 16.97m

<span style="font-family: Verdana,Geneva,sans-serif;">The hypotenuse is 16.97m.

__**Example 2:**__

Find the surface area of the cone.



The formula for the surface area of a cone is: SA = π//r//² + π//rs//
But, we are only given the height and radius, and not the side length. We can however, use the pythagoreom theorem to find the side length (hypotenuse), which will allow us to find the surface area. Thus:

A² + B² = C² 12² + 6² = C² 144 + 36 = C² 180 = C² √180 = C C = 13.41m S = 13.41m

We can then solve the original formula SA = π//r//² + π//rs//

SA = π//r//² + π//rs// SA //=// π//6//² + π//(6)(13.41)// SA //=// π36 + π80.46 SA = 113.1 + 252.77 SA = 365.87m²

<span style="font-family: Verdana,Geneva,sans-serif;">The surface area of the cone is 365.87m².

__**<span style="color: rgb(255, 130, 0); font-size: 120%;"> CHECK YOUR UNDERSTANDING (C):**__

<span style="color: rgb(239, 11, 11);">Determine the unknown measure of the shape below. Answers are located above "citations".



<span style="font-family: Verdana,Geneva,sans-serif;">__**Inside the Classroom and Beyond:**__ Optimal Values help us with many things, in and outside of school. Those who go on to work as architects, engineers, metalworkers, and other similar jobs will often deal with volume, surface area, perimeter and area every day. Even if you do not go on to work on one of these jobs, you might often find yourself dealing with optimal values anyways. For example, if your renovating your basement, you will need to measure and plan before building. What if you miscalculated the area, simply because you did not learn optimal values? Or if your fencing an area, and mistook the perimeter. In that case, you would probably spend extra money on unneeded fencing, or end up with too little fencing. Optimal Values are just as important outside the classroom.

Perimeter is used for many things in the world. They are used for many projects, seatwork and jobs. In school you use perimeter and area for labs and projects, along with homework and seatwork. In most elementary schools, there are whole units on perimeter and area, or Optimal Values. Outside of school, perimeter and area are used in a variety of jobs. People like architects, builders, and construction workers use perimeter and area everyday. For example, if a builder was designing a park he would need perimeter and area. He would need perimeter if he wanted to know the length of the boundaries or if he wanted to fence it. He would need area know how much room inside the park, if he wanted to put any equipment inside. There are many other uses for perimeter and area but there are also many uses for volume and surface area.
 * __Perimeter and Area:__**

Volume is an essential math tool used in our everyday lives. Without it, it would make our lives a lot harder and more difficult. It's taught from the very beginning of elementary school and is continued to be used all the way throughout highschool. Even outside of school, volume is used in our everyday lives. Without it, how would a manufactuer know how much room was left inside of a moving box? How would a juice company know how much room was left for the juice? They wouldn't, this is why volume is so important to us. There are just a few of many reasons of why we need volume, however, there are many other uses for volume and surface area.
 * __Volume and Surface Area:__**

__**Maximum Volume and Minimum Surface Area:**__

Finding the maximum volume and minimum surface area of 3D objects can be extremely useful. They allow us to maximize the use of the materials be used and create the most efficient shape. For example, if a company is packaging their product into boxes, using a cube shape would allow them to minimize the cardboard being used (surface area), and maximize the amount of space in that box (volume). The maximum volume might allow the company to package 1 or 2 more of their products into the box. By using the cube shape, the company will be saving money for each box being made. If the company was to make thousands of boxes, they would be saving quite a bit.

<span style="font-size: 170%; font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif; color: rgb(214, 46, 46);">__So Why Do We Care About Optimal Values?__

Most people do not realize how often every day jobs use them. Any job where building something is required, like a carpenter for example, needs to use perimeter and area to build and landscape someone's house. Laws and regulations require you to know how much space you will be taking up before you build objects such as houses, or pools, etc. Additionally, even when you're buying a house, chances are you are looking at how big it is-- the squared space. Most people do not think about how often everyday jobs use optimal values day to day.

__**Example # 1**__

When building a swimming pool, one would want to maximize the swimming area while minimizing the amount of excavating, concrete, and steel that you will have to pay for/buy. This requires you to calculate the surface area to build the walls of the pool. You would also want to know the volume of the pool, to know how much water will be required to fill it.

__**Example # 2**__

If you ever go into the business of farming, or if your family owns a farm, it is likely that there will be a point in time when you will have to build a fence. Since fences cost money for a given area, you would want to minimize the perimeter but maximize the area. Also on the topic of farming--building a barn, with extras such as stalls, or lofts. Once again, you want to minimize the amount of materials needed (perimeter), therefore the cost, given the amount of space you need in the barn.

<span style="font-family: 'Lucida Sans Unicode','Lucida Grande',sans-serif; font-size: 120%; color: rgb(0, 0, 255);">Need Additional Help? YouTube-Related Videos On This Topic:

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 * <span style="color: rgb(255, 130, 0); font-size: 120%;">__CHECK YOUR UNDERSTANDING__ ** answers

<span style="color: rgb(128, 0, 128);"> A. Perimeter is 260 cm, Area is 3525 squared cm. <span style="color: rgb(0, 128, 128);">B. Volume is 1809.56 cubed inches, Surface Area is 603.19 squared inches. <span style="color: rgb(215, 20, 20);">C. The unknown measure is 10 units.

<span style="font-family: Arial,Helvetica,sans-serif;">__Mathplease.com__. 5 June 2009 <www.mathplease.com>.
 * <span style="font-family: Verdana,Geneva,sans-serif;">//<span style="font-family: Arial,Helvetica,sans-serif;">Math Power 9 Ontario Edition // <span style="font-family: Arial,Helvetica,sans-serif;">. (1999). Toronto: McGraw-Hill Ryerson.
 * <span style="font-family: Arial,Helvetica,sans-serif;">Pierce, R. (2007, July 17). <span style="font-family: Verdana,Geneva,sans-serif;">//<span style="font-family: Arial,Helvetica,sans-serif;">Definition of Perimeter // <span style="font-family: Arial,Helvetica,sans-serif;">. Retrieved November 17, 2008, from http://www.mathsisfun.com/definitions/perimeter.html
 * <span style="font-family: Verdana,Geneva,sans-serif;">//<span style="font-family: Arial,Helvetica,sans-serif;">Yahoo! Answers // <span style="font-family: Arial,Helvetica,sans-serif;"> (2008). Retrieved December 17, 2008, from http://education.yahoo.com/homework_help/math_help/solutionimages/mini6and7gt/7/1/1/mini6and7gt_7_1_1_22_60/f-578-21-pr-q.gif
 * <span style="font-family: Arial,Helvetica,sans-serif;">Rancourt, S. (2009). MPM 1D1 - Course Notes
 * <span style="font-family: Arial,Helvetica,sans-serif;">Russell, D. (n.d.). //<span style="font-family: Arial,Helvetica,sans-serif;">Pythagorean Theorem //<span style="font-family: Arial,Helvetica,sans-serif;">. Retrieved June 5, 2009, from http://math.about.com/mbiopage.htm
 * <span style="font-family: Arial,Helvetica,sans-serif;">__Online Math Tutoring__. 2008. 5 June 2009 <http://www.tcyonline.com/etutoring >.

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