Question

A cardboard box without a lid is to have a volume of 5,324 cm3. Find the dimensions that minimize the amount of cardboard used. (Let x, y, and z be the dimensions of the cardboard box.)

Answer 1

The area of the box is

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

which we want to minimize subject to the constraint $xyz=5324$.

The Lagrangian is

$L(x,y,z)=xy+2xz+2yz+\lambda(xyz-5324)$

with critical points where the partial derivatives are 0:

$L_x=y+2z+\lambda yz=0$

$L_y=x+2z+\lambda xz=0$

$L_z=2x+2y+\lambda xy=0$

$L_\lambda=xyz-5324=0$

Notice that

$L_y-L_x=(x-y)+\lambda(xz-yz)=0\implies(x-y)(1+\lambda z)=0\implies x=y\text{ or }z=-\dfrac1\lambda$

Substituting the latter into either $L_x=0$ or $L_y=0$ will end up suggesting that $\lambda$ is infinite, so we throw out this case.

If $x=y$, then

$L_z=0\implies4x+\lambda x^2=0\implies x=0\text{ or }x=-\dfrac4\lambda$

We ignore the case where $x=0$ because that would make the volume 0. Then

$x=y=-\dfrac4\lambda\text{ and }L_x=0\implies-\dfrac4\lambda+2(1331\lambda^2)+\lambda\left(-\dfrac4\lambda\right)(1331\lambda^2)=0$

$\implies2662\lambda^3+4=0$

$\implies\lambda=-\dfrac{\sqrt[3]{2}}{11}$

so we have one critical point at

$(x,y,z)=\left(22\sqrt[3]{4},22\sqrt[3]{4},\dfrac{11}{2\sqrt[3]{2}}\right)\approx(34.9228,34.9228,4.3654)$

which give a minimum area of about 1829.41 sq. cm.