The area of the box is
[tex]A(x,y,z)=xy+2xz+2yz[/tex]
which we want to minimize subject to the constraint [tex]xyz=5324[/tex].
The Lagrangian is
[tex]L(x,y,z)=xy+2xz+2yz+\lambda(xyz-5324)[/tex]
with critical points where the partial derivatives are 0:
[tex]L_x=y+2z+\lambda yz=0[/tex]
[tex]L_y=x+2z+\lambda xz=0[/tex]
[tex]L_z=2x+2y+\lambda xy=0[/tex]
[tex]L_\lambda=xyz-5324=0[/tex]
Notice that
[tex]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[/tex]
Substituting the latter into either [tex]L_x=0[/tex] or [tex]L_y=0[/tex] will end up suggesting that [tex]\lambda[/tex] is infinite, so we throw out this case.
If [tex]x=y[/tex], then
[tex]L_z=0\implies4x+\lambda x^2=0\implies x=0\text{ or }x=-\dfrac4\lambda[/tex]
We ignore the case where [tex]x=0[/tex] because that would make the volume 0. Then
[tex]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[/tex]
[tex]\implies2662\lambda^3+4=0[/tex]
[tex]\implies\lambda=-\dfrac{\sqrt[3]{2}}{11}[/tex]
so we have one critical point at
[tex](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)[/tex]
which give a minimum area of about 1829.41 sq. cm.
