Question #e7bb3

1 Answer
Mar 9, 2017

#9 xx 6 xx 6#

Explanation:

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#