Maxima/Minima Problems PDF

# Maxima/Minima Problems 1. **Identify the constant** 2. **Identify the variable to be maximized or minimized.** 3. **Express the variable in terms of the other relevant variable to get a single variable equation.** 4. **Differentiate and equate to 0.** **SP** 1. What positive number added to its...

# Maxima/Minima Problems 1. **Identify the constant** 2. **Identify the variable to be maximized or minimized.** 3. **Express the variable in terms of the other relevant variable to get a single variable equation.** 4. **Differentiate and equate to 0.** **SP** 1. What positive number added to its reciprocal gives the minimum sum? Let: - x = required positive number - 1/x = reciprocal of the number - y = sum of x & ­1/x y = x + 1/x We get: y' = 1 - 1/x² = 0 y' = x² - 1 = (x+1)(x-1) Therefore, the number is 1. ## What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fence? - Area: A = xy - Perimeter: P = 2x + 2y Since A is constant, d(A)/dx = 0 d(A)/dx = xdy + y.dx --> xy' = -y y' = -y/x P = 2x + 2y dP/dx = 2*dx + 2*dy/dx 0 = 2 + 2(-y/x) 0 = 2 - 2y/x -2 = -2y/x -2x = -2y x = y Therefore, it is a square! ## A rectangular field of fixed area is to be enclosed and divided into 3 lots by parallels to one of the sides. What should be the relative dimensions of the field to make the amount of fencing minimum? - Area: xy - Area = Fixed A = xy 0 = d(y)/dx + x*d(y)/dx 0 = y + xy'; xy' = -y y' = -y/x - Perimeter: P = 2x + 4y P1 = 2 + 4y' = 0 0 = 2 + 4(-y/x) 4y = 2, 4y = 2x Dim: y = 2, x= 2 ## Find the dimensions of a rectangle with perimeter 1000 meters so that the area of the rectangle is a maximum. - Length = x (m) - Width = y (m) -Perimeter = 2x+2y - Area = xy (m²) 1000 = 2x+2y 500 = x+y y = 500-x A=xy x = 500 - x = 500x - x² A' = x(-1) + (500-x).1 =-x+500-x A' = 500-2x = 0 2x = 500 x = 250 m y = 500-250 = 250m Dim (250m x 250m) ## A square sheet of cardboard with each side "a" centimeter is to be used to make an open-top box by cutting a small square of cardboard from each of the corners, and bending up the sides. What is the side length of the small squares if the box is to have as large a volume as possible. - V = Lwh V= x (a - 2x) (a-2x) = 4x³ - 4ax² + a²x V' = 12x² - 8ax + a² (2x-a) (6x-a) = 0 x = a / 2 x = a / 6 y' = 24x - 8a Max Volume is attained when x = a/6 V = 4x³ - 4ax² + a²x Vmax = 4 (a/6)³ - 4a(a/6)² + a²(a/6) = 4a³ / 6³ - 4a³ / 6² + a³ / 6 = a³ (4 - 24 + 36) / 6³ = 16 a³ / 6³ = 4a³ / 27 = a³ * 4 / 27 V = 2a³ / 27 when side length = a / 6 ## 3 Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen? - Fencing: P = 2y + 5x - Fencing: F = 2Length + 5Width 500 = 5x + 2y 2y = 500 - 5x y = 250 - 5x A = LW = xy A = x ( 250 - 5x ) A = 250x - 5x² A' = 250 - 5 . 2x 250 - 5x = 0 5x = 250 x = 250/5 = 50 y = 250 - 5(50) = 125m Dim: (xy) = (50x125)m.