Question

Show that the surface area of a closed cuboid with square base and givenvolume is minimum, when it is a cube.

Answer 1

Let the base length be a and length be h.
$volume \: \: v = {a}^{2} h - - - (1) \\ surface \: area \: s = 2 {a}^{2} + 4ah$
substituting h from (1),
$s = 2 {a}^{2} + \frac{4v}{a} \\ \frac{ds}{da} = 4a - \frac{4v}{ {a}^{2} } - - - (2)\\ for \: the \: optimum \: \frac{ds}{da} = 0 \\ this \: gives \: {a}^{3 } = v \: - - - (3)$
$substituting \: v \: from \: \: (1) \: to \: (3)\\ {a}^{3} = {a}^{2} h \\ i.e. \: \: a = h \\ which \: means \: cuboid \: is \: a \: cube$
Differentiating (2) again,
$\frac{ {d}^{2}s }{d {a}^{2} } = 4 + \frac{8v}{ {a}^{3} } = 12 > 0 \\ thus \: (3) \: is \: minimum$

Answer 2

Hey !!

Let a closed cuboid have a base of x × x and a height y

Let its volume be V and surface area by S

V = x²y                    S = 2 (x² + 2xy)

  y = V / x²                 = 2x² + 4xy

                                  = 2x² + 4x. V / x²

                                   = 2x² + 4v / x

ds / dt = d / dx (2x² + 4v/x)

               = 4x + 4v (-1) / x²

                = 4x - 4v / x²

d²s / dx² = 4 - 4v(-2) / x³

              = 4 + 8v / x³

To maximise or minimise S , ds/dx = 0

  4x - 4v / x² = 0          d²s / dx² | x = v¹/³ = 4 + 8v / v = 12 which is > 0

     4x = 4v / x²                   Hence, S is minimum at x = v¹/³

     x³ = v¹/³

     x³ = v

     x³ = x²y

     x³ - x²y = 0

      x² (x-y) = 0

     x = y

Hence, the given cuboid is a cube of side x.

Good luck !!