Chapter 3: Many Variables; Section 1: Differentiation; page 6
Constraints, Exercise 2
Question
The volume of a straight circular cylinder is V0. Find the radius r of the base and the height h, that minimize the surface A of the cylinder?
By using the exclusion of one of the variables, with verification of the minimum!
By using the Lagrange multipliers! Compare the results!
Without minimization, what happens with the surface, for the limit of r→0 ?
Reminder
In some applications of functions of many variables, their domains are restricted by constraints. A constraint is a functional relation among the variables. If the constraint can be solved as the dependence of one variable by the others, then the substitution of this variable in the function reduces by one the number of variables of the original function. The new function obtained this way can have different stationary points.
(3.1.6.21)
... yields two equations
(3.1.6.25)
with three unknowns. By adding the constraint
(3.1.6.26)
there are three equations with three unknowns. Their solution fixes the values of x, y and λ, for the stationary points. The value of λ, called the Lagrange multiplier, is a bonus but does not contain any important information.
Parts 1-5
Solution of question 1
The function to be minimized is the surface area of the cylinder A :
and the constraint is the volume
The height h will be expressed by r, by using the constraint
and substituted in A
The stationary point corresponds to
which yields
and from part 2:
From part 3, we obtain
indicating a minimum.
Parts 6-9
Solution of question 2
From (3.1.6.21) and (3.1.6.25-26), and by use of part 1, the required equations for the solution of the question, by Lagrange multipliers are
The Lagrange multiplier λ can be excluded, by multiplying the first two equations by appropriate factors, and adding them up, according to:
yielding
The equations left for the solution are therefore
which should give exactly the same solution as part 4.
Indeed the substitution of h and r from part 1, in the equations of part 8 give
and
Part 10
Solution of question 3
In part 2, we substituted h from the constraint, into the surface S, and obtained the relation, where the surface is expressed only by the variable r:
and the limit is
Score
By parts.
Any one of the 10 parts is worth one point.
By questions.
Question 1 is worth 5 points.
Question 2 is worth 4 points.
Question 3 is worth one point.