Solve Box Cost Minimization w/ Lagrange Multipliers

[MarkFL]

Here is the question:

Please help!? lagrange multiplier method?


Find the dimensions of the box which will minimize the TOTAL COST of manufacturing the following open top box of volume 6ft^3?

Solve by the lagrange multiplier method!
Bottom panel costs $3/ft^2
side panel cost $.50/ft^2
Front and back panels cost $1/ft^2

I have posted a link there to this topic so the OP can view my work.

 

[MarkFL]

Hello acacia,

I would orient the box such that the width is $x$, the height is $y$ and the length is $z$. Hence, the bottom panel has area $xz$, the side panels have a total area of $2yz$ and the front and back panels have a combined area of $2xy$.

Let all linear measures be given in feet.

Thus, our objective function, the function we wish to minimize is the cost function in dollars, which is given by:

$$C(x,y,z)=3xz+yz+2xy$$

Subject to the constraint on the volume:

$$g(x,y,z)=xyz-6=0$$

Using Lagrange multipliers, we obtain:

$$3z+2y=\lambda(yz)$$

$$z+2x=\lambda(xz)$$

$$3x+y=\lambda(xy)$$

Solving for $\lambda$, the first two equations imply:

$$\frac{2y+3z}{yz}=\frac{2x+z}{xz}$$

Cross-multiplying, we obtain:

$$2xyz+3xz^2=2xyz+yz^2$$

$$3xz^2=yz^2$$

Since the constraint requires $$0<z$$, we may write:

$$3x=y$$

In like manner the first and third equations above imply:

$$\frac{2y+3z}{yz}=\frac{3x+y}{xy}$$

Cross-multiplying, we obtain:

$$2xy^2+3xyz=3xyz+y^2z$$

$$2x=z$$

Substituting for $y$ and $z$ into the constraint, we obtain:

$$x(3x)(2x)=6$$

$$x^3=1$$

$$x=1\implies y=3,\,z=2$$

Observing that:

$$C(1,3,2)=3(1)(2)+(3)(2)+2(1)(3)=18$$

and another constraint value such as $(x,y,z)=(1,2,3)$ yields:

$$C(1,2,3)=3(1)(3)+(2)(3)+2(1)(2)=19$$

We may then conclude:

$$C_{\min}=C(1,3,2)=18$$