Eagle The
Posted on Tuesday, 20 May, 2003 - 05:54 am:
Consider the cross section of a house:
House

a. Solve this problem by the method of Lagrange multipliers. Use the Pythagorean theorem,
a2 + b2 = c2 (*)
to eliminate one of the variables, thus reducing the problem to three variables with one constraint and one multiplier.

I got one as the answer for this part.

b. In part (a), the calculations were difficult because of the expressions involving radicals. To avoid these difficultites, solve this problem again by leaving (*) as a second constraint. Begin by extending the theory of Lagrange multipliers to allow for a second constraint and a second multiplier. Then solve the specific problem above using this idea.
I get stuck on this part, I get 5 equations but I cannot figure out how to solve them.
my system of equations:
0
=
lb + 2 ma
2
=
l(a + 2h) + 2 mb
2
=
m(-2c)
0
=
a b + 2h b - 30
0
=
a2 + b2 - c2

Thanks for any help
Andre Rzym
Posted on Tuesday, 20 May, 2003 - 08:20 am:
Could you clarify what the problem is please? a seems to be the vertical height of the roof, c the length of the roof (from ridge to gutter) and b the semi-width of the roof base.

So on the assumption that the roof is isosceles, it will be true that a2 +b2 =c2 , but what are you then trying to maximise/minimise?

Andre
Eagle The
Posted on Tuesday, 20 May, 2003 - 01:09 pm:
We want to minimize the perimeter subject to the area being 30 m2

You are correct about a, b, and c.
Andre Rzym
Posted on Tuesday, 20 May, 2003 - 04:08 pm:
Define h as height of the house (excluding the roof). We can state Pythagoras' theorem as
T = c2 -a2 -b2
where we require T=0. We can phrase the constraint on area as
A = 2bh + ab - 30
where we require A=0. We can write the perimeter as
P = 2(c+b+h)
So to minimise P subject to T=A=0, we define
Q = P + mT + lA
And require
0 = Q
a
 = Q
b
 = Q
c
 = Q
h
 = Q
l
 = Q
m
Giving rise to the equations
0
=
-2am+ b l
0
=
2 - 2b m+ l(2h+a)
0
=
2 + 2mc
0
=
2 + 2bl
0
=
2bh + ab - 30
0
=
c2 -a2 -b2
So
c
=
2a
b
=
  __
Ö3a
 
l
=
-l/   __
Ö3a
 
m
=
-l/2a
h
=
(30 -   ___
Ö3a2
 
 )/2   __
Ö3a
 
etc. Andre