MAXIMIZING AREA AND VOLUME

1. MAXIMIZING AREA AND VOLUME

2. EXAMPLE 1: The Starks have 60metres of fencing with which to make a rectangular dog run. If they use a side of the shed as one side of the run, what dimensions will give the maximum area? P = 2w + l A = lw shed A =w(60 – 2w) = 60w – 2w2 60 = 2w + l w w A′ = 60 – 4 w l = 60 – 2w 0 = 60 – 4 w l 4w = 60 w = 15 When w = 15, l = 60 – 2(15) = 30 The Starks should make their run 30 m by 15m to get a maximum area of 450m2 The second derivative is negative; thus, the area is a maximum. A′ = 60 – 4 w soA″= – 4

3. l w w w EXAMPLE 2: A rectangular field is to be enclosed and divided into 2 smaller plots by a fence parallel to one of the sides. Find the dimensions of the largest such field if 1200 m of fencing material is to be used. A = lw A = w(600– 1.5w) = 600w – 1.5w2 P = 3w+ 2l A′ = 600 – 3 w 1200 = 3w+ 2l 0 = 600 – 3 w 2l = 1200 – 3 w 3w = 600 l = 600 – 1.5w w = 200 When w = 200, l = 600 – 1.5(200) = 300

4. l w w w A′ = 600 – 3 w The second derivative is negative; thus, the area is a maximum. A″= - 3 The dimensions of the field should be 300 m by 200m to get a maximum area of 60 000m2

5. x x y EXAMPLE 3: A box has square ends and the sides are congruent rectangles. The total area of the four sides and two ends is 294 cm2. What are the dimensions of the box if the volume is a maximum and what is the maximum volume? volume = lxwx h V = x2 y surface area = 2x2 + 4xy 294 = 2x2 + 4xy 294 – 2x2 = 4xy

6. When x = 7 cm V′ = 73.5 – 1.5x2 73.5 – 1.5x2 = 0 1.5x2 = 73.5 Box is 7cm x 7cm x 7 cm Max Vol = 343 cm3 x2 = 49 x = 7 V″ = –3x= 3(7) = – 21 The second derivative is negative; thus, the volume is a maximum.

7. x x y EXAMPLE 4 An open (it has no top) box has square ends and the sides are congruent rectangles. The total area of the four sides and one end is 192 cm2. What are the dimensions of the box if the volume is a maximum and what is the maximum volume? surface area = x2 + 4xy volume = lxwx h 192 = x2 + 4xy

8. V′ = 48 – 0.75x2 48 – 0.75x2 = 0 0.75x2 = 48 Box is 8cm x 8cm x 4 cm Max Vol = 256 cm3 x2 = 64 x= 8 V″ = –1.5x= –1.5(8) = –12 The second derivative is negative; thus, the volume is a maximum.

9. 3 y 2 2 3 x EXAMPLE 5 A page contains 600 cm2. The margins at the top and bottom are 3 cm. The margins at each side are to be 2 cm. What are the dimensions of the paper if the printed area is a maximum? Dimensions of printed area are (x – 4) by (y – 6) Area of page: x y = 600 A = (x – 4)(y – 6) A = (x – 4)( 600 x- 1 – 6) A = 600 – 6x – 2400x-1 + 24 A = 624 – 6x – 2400 x-1

10. A = 624 – 6x – 2400 x-1 When x = 20, The dimensions of the page are 20 cm by 30 cm. x2 = 400 x = 20 The second derivative is negative when x = 20; thus, the area is a maximum.

11. x 18 – 2x 18 cm 48 – 2x 48 cm EXAMPLE 6 A rectangular open-topped box is to be made from a piece of material 18 cm by 48 cm by cutting a square from each corner and turning up the sides. What size squares must be removed to maximize the capacity of the box? V′ = 12x2 – 264x + 864 V = lwh = x(18 – 2x)(48 – 2x) 12x2 – 264x + 864 = 0 V = x (864 – 132x + 4x2) x2 – 22x + 72 = 0 V = 864x – 132x2 + 4x 3 (x – 18)(x – 4) = 0 x = 18 or x = 4 The squares should be 4 cm by 4 cm to create a maximum volume of 4 x 40 x 10 = 1600 cm3