Noah Brown John Riley Justine Bailey Solving Applied

Noah Brown John Riley Justine Bailey Solving Applied Max/Min Problems

RIDDLE � 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”?

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

What is a Max/Min Problem? �A problem to which the answer is the Maximum or Minimum quantity of a specified measurement using a certain amount of a different quantity

Step 1 �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

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

Step 3 �Set the equation to equal a single variable �Example: 500 ft = 2 X + 2 Y �(500 ft – 2 X) / 2 = Y �Y = -X + 250

Step 4 �Substitute “Y” in the second equation for the equation found on the previous slide �Example: A = XY �A = X(-X + 250) �A = -X 2 + 250 X

Step 5 �Take the derivative of the equation on the previous slide �Example: A = -X 2 + 250 X �d. A/d. X = -2 X + 250

Step 6 �Find the zeros of the derivative, either by graphing or by equation �Example: d. A/d. X = -2 X + 250 � 0 = -2 X + 250 �X = 125

Step 7 �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: + �|----------| 0 125 250 125 is a maximum

Step 8 �Plug in the maximum “X” value into the original equation to find “Y” �Example: � 500 ft = 2(125) + 2 Y � 250 ft = 2 Y � 125 ft = Y

Step 9 �Plug the “X” and “Y” values into the area/volume equation �Example: Area = XY Area = (125) Area = 15, 625 ft 2

EXAMPLE PROBLEM #1 � 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?

Problem #1 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

Example Problem #2 �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?

Problem #2 � 5π=πr² + (2πr)h Volume= πr² ((5/2 r)-(r/2)) (5/2)πr-(1/2)πr³ �Differentiate V’ = (5/2)π-(1/2)3πr² =(1/2)π(5 -3 r²) 0=(1/2)π(5 -3 r²) r=√(-5/3)

Sources �"Calculus AB: Applications of the Derivative. " Spark. Notes, 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.