Given Volume:

$V = x \, (3x) \, y

V = 3x^2 \, y$

$0 = 3x^2 \, y' + 6xy$

$y' = -2y/x$

Total Area:

$A_T = 2(3x^2) + 2(3xy) + 2(xy)$

$A_T = 6x^2 + 8xy$

$dA_T / dx = 12x + 8 \, (x y' + y) = 0$

$12x + 8 \, [ \, x (-2y/x) + y \, ] = 0$

$12x + 8 \, [ \, -2y + y \, ] = 0$

$12x = 8y$

$y = \frac{3}{2} x$

Altitude = 3/2 × shorter side of base. *answer*