Calling #a,b,c# the dimensions we have a restriction
#a b c = 324#
and an objective function
#f(a,b,c) = c_1(2(ab+ac)+bc)+c_2bc#
where #c_1,c_2# are the construction costs with the relationship
#c_2 = 2c_1# or
#f(a,b,c) = c_1((2(ab+ac)+bc)+2bc)#
This problem can be handled using the so called Lagrange multipliers
The lagrangian
#L(a,b,c,lambda) = f(a,b,c)+lambda(abc-324)#
The stationary points are the solutions to
#grad L = vec 0#
where #grad=(partial/(partial a),partial/(partial b),partial/(partial c),partial/(partial lambda))# so we have
#(L_a,L_b,L_c,L_(lambda))=(0,0,0,0)# or
#{(2 (b + c) c_1 + b c lambda=0),( (2 a + 3 c) c_1 +
a c lambda=0), ((2 a + 3 b) c_1 + a b lambda=0), ( a b c-324=0):}#
Solving for #a,b,c,lambda# we obtain
#a=9,b=6,c=6,lambda=-2c_1/3#
