Question

find the dimensions of the rectangular box with largest volume if the total surface area is given as 16 cm2. (let x, y, and z be the dimensions of the rectangular box.)

Answer 1

With dimensions x-by-y-by-z, the volume of such a box is

`V(x,y,z)=xyz`

and its surface area is

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

Given that A = 16 cm², we can solve for z in the constraint:

`2 (xy + xz + yz) = 16 \implies z = (8-xy)/(x+y)`

Substitute this into the volume function to reduce it to a function of 2 variables:

`V\left(x,y,(8-xy)/(x+y)\right) = V^*(x,y) = ((8-xy)xy)/(x+y)`

Find the critical points of V*, where both its partial derivatives vanish:

`(\partial V^*)/(\partial x) = (y^2(8-x^2-2xy))/((x+y)^2) = 0`

`(\partial V^*)/(\partial y) = (x^2(8-y^2-2xy))/((x+y)^2) = 0`

There's no box if either x = 0 or y = 0, so the critical points occur when

`\begin{cases}8-x^2-2xy = 0 \\ 8 - y^2 - 2xy = 0\end{cases}`

By combining the equations, we find

`(8-y^2-2xy)-(8-x^2-2xy)=0-0 \implies x^2-y^2=0 \implies |x|=|y|`

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

`x=y`

Substitute this into V* to once again reduce the number of variables:

`V^*(x,x) = V^(**)(x) = ((8-x^2)x^2)/(2x) = 4x - \frac{x^3}2`

Find the critical points of V** :

`(dV^(**))/(dx) = 4 - \frac{3x^2}2 = 0 \implies x = \pm 2√(\frac23)`

but again, x must be positive.

Putting everything together, there is only one critical point at

`(x,y,z) = \left(2√(\frac23), 2√(\frac23), 2√(\frac23)\right)`

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.)