1. Optimizing Area/SA/VolumeC2
Optimizing Area/SA/Volume

2. Area
To optimize means to find the conditions needed to give the most area for a given length of materials.
In designing rectangular enclosures, we use the A = L(w) formula to test it out.
Ex) You have 16 m of fence, and a garden needs to be closed on 4 sides. What is the maximum area you can get?

3. Area
Solution: You need to draw out the possible cases and calculate the areas using 16 m:
Case 1: 1m x 7m = 7 m2
Case 2: 2m x 6m = 12 m2
Case 3: 3m x 5m = 15 m2
Case 4: 4m x 4m = 16 m2
Case 4 yielding a square gives the most area if all 4 sides are used.

4. Area
Ex2) You have 16 m of fence, and a garden needs to be closed on 3 sides. What is the maximum area you can get?
Case 1: 1m x 14m = 14 m2
Case 2: 2m x 12m = 24 m2
Case 3: 3m x 10m = 30 m2
Case 4: 4m x 8m = 32 m2
Case 5: 5m x 6m = 30 m2
Case 4 gave the most area.
Trial and error should be used.

5. Minimize the surface area of a Square-Based Prism
Boxes used in packaging companies must be appealing, suitable for the product and cost efficient.
Popcorn can be given in a square based prism box. Given a volume of cm3, what dimensions will minimize the cardboard used?
A cube uses the least SA for a given volume.
V = s3
when all sides are equal!
= s3
Cube root both sides:
s = 79.4 cm

6. Maximize the volume of a cylinder for a given amount of material
To maximize volume, pop can companies want to use the least amount of metal to save costs.
Cylinder SA is 2πr2 + 2πrh and V = πr2h
Ex1) A can has to be made with 600 cm2 of metal. Determine the dimensions that yield maximum volume.
Condition to do this is when the diameter equals the height!! Sub in h = 2r into SA formula:
SA = 2πr2 + 2πrh
= 2πr2 + 2πr(2r)
= 2πr2 + 4πr2
= 6πr2

7. Cylinders continued
Use the metal area to find the cylinder dimensions: 600 = 6πr2 100 = πr2 100/π = r = r2 Square root both sides r = 5.64 cm The height should be twice this so h = 2(5.64 cm) = cm The max Volume of the can is: V = π(5.64 cm)2(11.28 cm) = 1127 cm3

8. Cylinders again
A can must hold 1000 cm3 of a liquid. (1 L). What dimensions will minimize the material used?
Sub in h = 2r into Volume formula:
V = πr2h
V = πr2(2r)
= 2πr3
Now solve for r:
1000 = 2πr3
500/π = r3
159 = r3
r = 5.42 cm

9. Cylinders again
Now find the SA h = 2r = 2(5.42 cm) = cm: SA = 2πr2 + 2πrh = 2π(5.42 cm)2 + 2π(5.42 cm)(10.84 cm) ≈ 554 cm2

10. Summary
Optimizing area of a rectangle is done when a square is used, given a fixed perimeter.
If fencing: Max area depends on number of sides enclosed (4 sides is still a square).
For a square-based prism of given volume, a cube gives minimum surface area (Use V = s3)
For a square-based prism of given surface area, a cube gives maximum volume (Use SA = 6s2)
For a cylinder of given volume, 2r = h gives the minimum surface area (Use V = 2πr3)
For a cylinder of given surface area, 2r = h gives the maximum volume (Use SA = 6πr2)