Mathematics G 12 Project
Maximizing Volume
A Real-World Optimization Challenge
Maximizing Volume
Let the length of the rectangular base be x and the height be h.
Since the base is squared, both length and width will be x.
108 − x 2
h=
4x
108 − x 2
108 𝑥 − x 3
108 𝑥 − x 3
V=x .
=
=
4x
4
4
108 𝑥
1 3
1 3
=
̶ x = 27x ̶ x
4
4
4
2
Maximizing Volume
3. Set the derivative equal to zero to find critical points.
27 ̶
3 2
x = 0
4
3 2
x = 27
4
3x2 = 108
x2 = 36
x = 6 cm
( can’t be ̶ 6 )
Maximizing Volume
4. Look at endpoints (if there are any).
Feasible domain : 0 ≤ x ≤ 108
5. If there are endpoints, plug critical numbers and endpoints
into the original equation to determine max.
V (0) = 0
V ( 108 ) = 27 108 ̶
V (6) = 27 (6) ̶
1
(27x
4
1
(6)2 = 108
4
̶
max
1
(
4
108)2 = 0
Maximizing Volume
6. If no endpoints use 1st or 2nd Derivative Test.
−3
V”(x) =
x= 0
2
−3
V”(6) =
(6) =
2
at x = 6
̶9˂0
33
Rel max
6
Then Substitute in h :
6
108 − x 2
108 − 62
108 − 36
72
h =
=
=
=
= 3
4x
4(6)
24
24
So, maximum volume occurs with dimensions 6 × 6 × 3.
6
6
Thank You