With dimensions x-by-y-by-z, the volume of such a box is
[tex]V(x,y,z)=xyz[/tex]
and its surface area is
[tex]A(x, y, z) = 2 (xy + xz + yz)[/tex]
Given that A = 16 cm², we can solve for z in the constraint:
[tex]2 (xy + xz + yz) = 16 \implies z = \dfrac{8-xy}{x+y}[/tex]
Substitute this into the volume function to reduce it to a function of 2 variables:
[tex]V\left(x,y,\dfrac{8-xy}{x+y}\right) = V^*(x,y) = \dfrac{(8-xy)xy}{x+y}[/tex]
Find the critical points of V*, where both its partial derivatives vanish:
[tex]\dfrac{\partial V^*}{\partial x} = \dfrac{y^2(8-x^2-2xy)}{(x+y)^2} = 0[/tex]
[tex]\dfrac{\partial V^*}{\partial y} = \dfrac{x^2(8-y^2-2xy)}{(x+y)^2} = 0[/tex]
There's no box if either x = 0 or y = 0, so the critical points occur when
[tex]\begin{cases}8-x^2-2xy = 0 \\ 8 - y^2 - 2xy = 0\end{cases}[/tex]
By combining the equations, we find
[tex](8-y^2-2xy)-(8-x^2-2xy)=0-0 \implies x^2-y^2=0 \implies |x|=|y|[/tex]
since √(x²) = |x| for any real x. But there's also no box if either x < 0 or y < 0, so by definition of absolute value, this condition reduces to
[tex]x=y[/tex]
Substitute this into V* to once again reduce the number of variables:
[tex]V^*(x,x) = V^{**}(x) = \dfrac{(8-x^2)x^2}{2x} = 4x - \dfrac{x^3}2[/tex]
Find the critical points of V** :
[tex]\dfrac{dV^{**}}{dx} = 4 - \dfrac{3x^2}2 = 0 \implies x = \pm 2\sqrt{\dfrac23}[/tex]
but again, x must be positive.
Putting everything together, there is only one critical point at
[tex](x,y,z) = \left(2\sqrt{\dfrac23}, 2\sqrt{\dfrac23}, 2\sqrt{\dfrac23}\right)[/tex]
so the box with maximum area is a cube with side length 2√(2/3) ≈ 1.633 cm, with volume 16/3 √(2/3) ≈ 4.3547 cm³.
(Note that this problem is also amenable to using Lagrange multipliers, which might even save some work.)
