1. Noah Brown John Riley Justine Bailey

2.  You are given 16 feet of flexible fencing in which to acquire “property”.  The fence must form a self contained shape  What is the optimum amount of area you can claim for yourself as “property”?

3.  The answer is infinity!  Make a square of it around yourself, and claim yourself to be on the outside!

4.  A problem to which the answer is the Maximum or Minimum quantity of a specified measurement using a certain amount of a different quantity

5.  Determine the quantity that is to be minimized or maximized  This will generally be area or volume  Example: Find the maximum area 500 ft of fencing can make into a rectangular pen

6.  Find a mathematical expression capable of determining the quantity, and any other expression deemed necessary  Example: 500ft = 2X + 2Y  Area = XY 0 ≤ X ≤ 500

7.  Set the equation to equal a single variable  Example: 500ft = 2X + 2Y  (500ft – 2X) / 2 = Y  Y = -X + 250

8.  Substitute “Y” in the second equation for the equation found on the previous slide  Example: A = XY  A = X(-X + 250)  A = -X 2 + 250X

9.  Take the derivative of the equation on the previous slide  Example: A = -X 2 + 250X  dA/dX = -2X + 250

10.  Find the zeros of the derivative, either by graphing or by equation  Example: dA/dX = -2X + 250  0 = -2X + 250  X = 125

11.  Make a sign chart for X to determine if the graph is concave up or down Up = Maximum, positive to negative Down = Minimum, negative to positive  Example: + -  |----------|----------|125 is a maximum 0 125 250

12.  Plug in the maximum “X” value into the original equation to find “Y”  Example:  500ft = 2(125) + 2Y  250ft = 2Y  125ft = Y

13.  Plug the “X” and “Y” values into the area/volume equation  Example: Area = XY Area = (125)(125) Area = 15,625 ft 2

14.  Problem : You decide to walk from point A (see figure below) to point C. To the south of the road through BC, the terrain is difficult and you can only walk at 3 km/hr. Along the road BC you can walk at 5 km/hr. The distance from point A to the road is 5 km. The distance from B to C is 10 km. What path you have to follow in order to arrive at point C in the shortest ( minimum ) time possible?

15. Step 1: Quadratic Formula 5²+x²=c² Step 2: (AB distance)/(AB rate) + (BC Distance)/(BC rate) =Time T=(√(25-x²)/3) + (√(15-x²)/5) Step 3: Take Derivative Step 4: Set T’=O Step 5: Create a Sign Chart

16.  A cylinder has no top and a surface area of 5π ft². What height and radius of the base will allow for the maximum volume of the Cylinder?

17.  5π=πr² + (2πr)h Volume= πr² ((5/2r)-(r/2)) (5/2)πr-(1/2)πr³  Differentiate V’ = (5/2)π-(1/2)3πr² =(1/2)π(5-3r²) 0=(1/2)π(5-3r²) r=√(-5/3)

18.  "Calculus AB: Applications of the Derivative." SparkNotes. SparkNotes, n.d. Web. 17 May 2013.  "Min, Max, Critical Points." Min, Max, Critical Points. N.p., n.d. Web. 17 May 2013.  "MAXIMUM AND MINIMUMVALUES." Maximum and Minimum Values. N.p., n.d. Web. 17 May 2013.  "Maximum/Minimum Problems." Maximum/Minimum Problems. N.p., n.d. Web. 17 May 2013.