Open Rectangular Box using Lagrange Multiplier

Question

Q: A rectangular container is open at the top and must have a volume of 10 m³. The material for the sides costs C dollars per m², while the material for the bottom costs 2C dollars per m². Find the optimal dimensions using Lagrange multipliers so as to minimize total cost of the container and what are the units for λ?

I am having trouble approaching the problem, only knowing that we start off with V= Lwh = 10 for the volume of the box. While cost = 2(xz) + 1(2xy + 2yz).

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My approach (I'm not sure if it's the right answer)

I found out that my x and y and z are 2.154 m. Is this correct?

My 2c crosses out when I isolate

I get 2cxz² = 2cyz²
2cxy²= 2cy²z
2cx²y = 2cx²z

x = y
x=z
y=z thus with these 3 we get that
y=z=x

plugging it into xyz = 10 we get x³ = 10 so x approximately equals to 2.154 m as well as y and z. So therefore the dimensions overall for the rectangular box at minimum cost is 2.154 x 2.154 x 2.514?

How do I find lambda units? Does this make sense?

How do I find lambda units or lambda itself?

Answer 1

So for lagrange multipliers, the formula used is:

$$ \nabla f = \lambda\nabla g $$

where $f$ is the function that needs to be optimized, $g$ is the constraint function, $\nabla$ is the gradient operator, and $\lambda$ is the lagrange multiplier.

So for for a open box, Let the volume be

$$V(l,w,h)=lwh=10$$

and since the bottom cost twice as much as the sides, let the cost be $$C(l,w,h)= 2clw + 2clh +2cwh$$

where $c$ was the cost of the material.

So $C$ is the function that is need to be minimized and $V$ is the constraint. Therefore, our lagrange formula would be

$$\nabla C(l,w,h,)=\lambda\nabla V(l,w,h)$$

by substituting in the equations, $$\nabla (2clw + 2clh +2cwh)=\lambda\nabla (lwh)$$

and then doing the gradient operation,

$$ <2cw+2ch,2cl+2ch,2cl+2cw> = \lambda<wh,lh,lw>$$

and then splitting into equations for each component,

$$ 2cw+2ch =\lambda wh \\ 2cl+2ch =\lambda lh \\ 2cw+2cl = \lambda lw $$

along with our constraint,

$$lwh=10$$

gives us a system of four equations with four unknowns: $l,w,h,\text{and } \lambda$.

then you can solve this system using any method you prefer.

You then find out that $l=w=h= \sqrt[3]{10}$

The dimensions of your box would be $\sqrt[3]{10} \times \sqrt[3]{10} \times \sqrt[3]{10}$.

As for your second question of what units $\lambda$ is, it has to be of units $\frac{\$}{m}$ in order for the units to match up in the system of equations.