Minima Maxima: 9a³y=x(4a-x)³

I think something is wrong with your given. If you check for x = a, y ≠ 3a. Here is the checking of your answer key.

Set x = a and solve for y
$9a^2 y = a(4a + a)^3$

$9a^2 y = 125a^4$

$y = \dfrac{125a^2}{9}$   ←   not equal to 3a
 

Kindly check the given.
 

$9a^3 y = x(4a - x)^3$   Differentiate y in terms of x using product formula $9a^3 y' = -3x(4a - x)^2 + (4a - x)^3$   Simplify (optional) $9a^3 y' = (4a - x)^2 [-3x + (4a - x)]$ $9a^3 y' = (4a - x)^2 (4a - 4x)$ $9a^3 y' = 4(4a - x)^2 (a - x)$   Determine the 2nd derivative $9a^3 y'' = -4(4a - x)^2 - 8(4a - x)(a - x)$ $9a^3 y'' = -4(4a - x)[ (4a - x) + 2(a - x) ]$ $9a^3 y'' = -4(4a - x)(6a - 3x)$ $9a^3 y'' = -12(4a - x)(2a - x)$   Set y' to zero for maxima and/or minima $0 = 4(4a - x)^2 (a - x)$   For (4a - x)2 = 0:

$x = 4a$   $9a^3 y = 4a(4a - 4a)^3$ $y = 0$   Check if Maxima or Minima. $9a^3 y'' = -12(4a - 4a)(2a - 4a)$ $y'' = 0$   ←   inflection point (the curve is neither upward nor downward)   Hence, (4a, 0) is neither maximum nor minimum

  For a - x = 0

$x = a$   $9a^3 y = a(4a - a)^3$ $y = 3a$   Check if Maxima or Minima. $9a^3 y'' = -12(4a - a)(2a - a)$ $y'' = (-)$   ←   the curve is concave downward   Hence, (a, 0) is maximum.

  Here is the graph of the curve with a = 1