# Question e7bb3

Mar 9, 2017

$9 \times 6 \times 6$

#### Explanation:

Calling $a , b , c$ the dimensions we have a restriction

$a b c = 324$

and an objective function

$f \left(a , b , c\right) = {c}_{1} \left(2 \left(a b + a c\right) + b c\right) + {c}_{2} b c$

where ${c}_{1} , {c}_{2}$ are the construction costs with the relationship

${c}_{2} = 2 {c}_{1}$ or

$f \left(a , b , c\right) = {c}_{1} \left(\left(2 \left(a b + a c\right) + b c\right) + 2 b c\right)$

This problem can be handled using the so called Lagrange multipliers

The lagrangian

$L \left(a , b , c , \lambda\right) = f \left(a , b , c\right) + \lambda \left(a b c - 324\right)$

The stationary points are the solutions to

$\nabla L = \vec{0}$

where $\nabla = \left(\frac{\partial}{\partial a} , \frac{\partial}{\partial b} , \frac{\partial}{\partial c} , \frac{\partial}{\partial \lambda}\right)$ so we have

$\left({L}_{a} , {L}_{b} , {L}_{c} , {L}_{\lambda}\right) = \left(0 , 0 , 0 , 0\right)$ 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

$a = 9 , b = 6 , c = 6 , \lambda = - 2 {c}_{1} / 3$