A25 & 26-Optimization (max & min problems)

Guidelines for Solving Applied Minimum and Maximum Problems 1.Identify all given quantities and quantities to be determined. Define your variables. If possible, make a sketch. 2.Determine the restrictions on the variables. 3.Find a secondary equation that relates to the problem. 4.Write a primary function for the quantity that is to be maximized or minimized. (If you need area and volume formulas, they are inside the back cover.) 5.Use the primary and secondary equations to create a function where you plug the secondary function into the primary function. 6.Take the derivative of the function, set it equal to zero, solve the equation to find the critical points, and determine where the relative maximum or minimum is.

Find two positive numbers whose sum is 20 and whose product is as large as possible. Define Variables: Let x = first number Let y = second number Restrictions: 0 < x < 20, 0 < y < 20 Write secondary equation: x + y = 20 So, y = 20 – x Write function: f(x) = xy f(x) = x(20 – x) = 20x – x 2 Find extrema (take derivative, set equal to zero, solve then interval test) f ’ (x) = 20–2x=0 => x=10 f ’ (x) + – f ’ (x) changes from + to – at x=10 so Rel. max at x = 10 Answer the question: The two numbers are 10 and 10.

Find two positive numbers that minimize the sum of twice the first number plus the second if the product of the two numbers is 288. Define Variables: Let x = first number Let y = second number Restrictions: x > 0, y > 0 Write secondary equation: xy = 288 So, y = 288/x Write function: f(x) = 2x+y Find extrema (take derivative, set equal to zero, solve then interval test) x = 12 f ’ (x) – f ’ (x) changes from – to + at x=12 so Rel. min at x = 12 y=288/12 = 24 Answer the question: The two numbers are 12 and 24.

A gardener has 200 ft. of fencing to enclose a rectangular planting bed. What length and width should it have so that its area is maximum? Draw a picture: Define Variables: Let x = length Let y = width Restrictions: x > 0, y > 0 Write secondary equation: 2x + 2y = 200 So, y = 100 – x Write function: f(x) = xy f(x) = x(100–x) = 100x – x 2 Find extrema (take derivative, set equal to zero, solve then interval test) f ’ (x) = 100–2x=0 => x = 50 f ’ (x) + – 0 50 f ’ (x) changes from + to – at x=50 so Rel. max at x =50 y = 100 –50 = 50 Answer the question: The length and width are 50 ft and 50 ft. x x y y

Find extrema (take derivative, set equal to zero, solve then interval test) f ’ (x) = 3 – 2x = 0 => f ’ (x) + – f ’ (x) changes from + to – at x=1.5 so Rel. max at x = 1.5 y = 3 – 1.5 = 1.5 Answer the question: the dimensions are 1.5 units by 1.5 units & area=2.25units 2 The upper right vertex of a rectangle lies on the line y = 3 – x. The left side of the rectangle is on the y-axis and the bottom side of the rectangle is on the x-axis. What is the largest area the rectangle can have, and what are its dimensions? Write Secondary Equation Write Function: x y (x,y) Draw a picture: Define Variables Restrictions: 0<x<3 and 0<y<3

A rectangle is bounded by the x-axis and the semicircle y = √(4-x 2 ) What length and width should the rectangle have so that its area is a maximum? Draw a picture: Define Variables: 2x = Length; y = width Restriction: 0<x<2, 0<y<2 Write secondary equation: y = √(4-x 2 ) Write function: f(x) = 2xy f(x)= 2x√(4-x 2 ) =2x(4-x 2 ) ½. Answer:length & width (x,y) x y Find extrema (take derivative, set equal to zero, solve then interval test) f ’ (x) + – f ’ (x) changes from + to – at x=√2 so Rel. max at x = √2

End of Day 1

A rectangular page is to contain 24 in 2 of print. The margins at the top and bottom of the page are each 1½ in wide. The margins on each side are 1 in. What should the dimensions of the page be so that the least amount of paper is used? Draw picture: Define Variables: Let x = width Let y = height Restrictions: x > 0, y > 0 Write secondary equation: xy = 24 So, y = 24/x Write f(x): f(x) =(x+2)(y+3) f(x) = (x+2)((24/x)+3) =24+3x+48x Find extrema (take derivative, set equal to zero, solve then interval test) f ’ (x) = 3-48x -2 = 0 => x = 4 f ’ (x) – f ’ (x) changes from – to + at x=4 so Rel. min at x=4 so x+2=6 y = 24/4 = 6 so y+3 = 9 Answer the question: The dimensions of the page: 6 inches by 9 inches. x x+2 y+3 y

You are planning to make an open rectangular box from an 8-in by 15-in piece of cardboard by cutting congruent squares from the corners & folding up the sides. What are the dimensions of the box of largest volume made this way, & what is its volume? Draw picture: Define Variables: Let x=side of cut square Restrictions: 0 < x < 4 Write f(x): Volume = f(x) =x(8-2x)(15-2x) f(x) = (8x-2x 2 )(15-2x) Find extrema (take derivative): f ’ (x) = (8x-2x 2 )(-2)+(15-2x)(8-4x) f ’ (x) =12x 2 -92x+120 f ’ (x) =4(3x-5)(x-6) Find extrema ( set deriv. equal to zero, solve then interval test) f ’ (x)=4(3x-5)(x-6)=0 => x = 6,x=5/3 f ’ (x) + – 0 5/3 4 f ’ (x) changes from + to – at x=5/3 so Rel. max at x=5/3 Answer the question: The dimensions of the page:14/3 in. by 35/3 in. by 5/3 in. w/ vol = 2450/27in 3 x 8-2x 15-2x x x x