Computes the volume, V (m³),
of a cylinder, constrained to an amount available, C ($),
for the areas of the basis, k = 1, or with top, k = 2,
and lateral surface.
The specific cost ($ ⁄ m²) for the basis is cB
and for the lateral surface is cL.
The functions for the volume and for the amount available (constraint) are:
V = π (D² ⁄ 4) h
C = cL π D h +
cB k π (D² ⁄ 4)
For the classical problem, C ≡ A,
total area, and, e.g.,
cB = cL = 1, giving:
for k = 2, D* = h*; and
for k = 1, D* = 2 h*.
The optimum values come from substituting h
(easier than D), from the constraint, into V, and differentiating.
It becomes:
D* = 2 √ [C ⁄
(3 π k cB]
h* = (1 ⁄ cL)
√ [C k cB ⁄ (3 π)]
Notice that these expression are
dimensionally homogeneous.
Plots V and h vs. D.
Other suggested data: k = bottom only . |
