V = ½x(n² - 4nx + 4x²)
= ½n²x - 2nx² + 2x³

dV/dx = ½n² - 4nx + 6x²

For stationary value of V, dV/dx = 0

½n² - 4nx + 6x² = 0

12x² - 8nx + n² = 0

Using quadratic formula,

x = ( -b ± √(b² - 4ac) ) / 2a

x = ( -(-8n) ± √((-8n)² - 4(12)(n²)) ) / 2(12)

x = ( 8n ± √(64n² - 48n²) ) / 24

x = ( 8n ± √(16n²) ) / 24

x = (8n ± 4n) / 24

(we can square root 16n² to get 4n. No need to worry about ± sign here as n is positive, so 16n² is also positive and the square root of a positive is represented by the principal root. eg √4 = 2 and not -√2 )

x = ⅓ n ± 1/6 n

x = ½n or x = 1/6 n

Now,

When x = ½n,
2x = n

but if n = 2x, then n - 2x = 0
This suggests that the length of the box is 0, which is not possible.

The same argument goes for the width of the box.

So x = ½n is rejected.

Therefore x = 1/6 n

dV/dx = ½n² - 4nx + 6x²

So d²V/dx² = -4n + 12x

= 12(1/6 n) - 4n
= 2n - 4n
= -2n < 0 (maximum)

(since n > 0, which means -2n < 0)

So the maximum value of V is obtained when x = 1/6 n

V = ½n²x - 2nx² + 2x³

= ½n²(1/6 n) - 2n(1/6 n)² + 2(1/6 n)³

= 1/12 n³ - 1/18 n³ + 1/108 n³

= 9/108 n³ - 6/108 n³ + 1/108 n³

= 4/108 n³

= 1/27 n³