Calculating Variable from a Derivative

[odolwa99]

My answer for part (i) matches up with that of the book but solving for part (ii) has gone awry. Can anyone help me clear this up?
Many thanks.

Homework Statement

Q. The given diagram (see attachment) shows part of the graph of y = x². p(x,y) is a point on the curve & a(6,0). bp $\perp$ ab. (i) Express the coordinates of p in terms of x only, (ii) Find the value of x if the area of the triangle abp is a maximum & hence find this maximum area.

Homework Equations

The Attempt at a Solution

(i) If y = x² => p(x,y) => p(x,x²)

(ii) $\frac{1}{2}$[(y₂ - y₁)(x₁ - x₃) - (x₂ - x₁)(y₁ - y₃)] => $\frac{1}{2}$[(0 - x²)(x - x) - (6 - x)(x² - 0)] => $\frac{1}{2}$[0 - (6x² - x³)] => $\frac{1}{2}$[-6x² + x³]
$\frac{dA}{dx}$ = $\frac{1}{2}$(-12x + 3x²) = 0 => -12x + 3x² = 0 => 3x² = 12x => x² = 4x => x = 4
$\frac{d^2A}{dx^2}$ = $\frac{1}{2}$(-12 + 6x) => $\frac{1}{2}$(-12 + 6(4)) => $\frac{1}{2}$ => $\frac{1}{2}$(12) => 6 > 0, making x = 4 a minimum.
Area = $\frac{1}{2}$(-6x² + x³) => $\frac{1}{2}$(-6(4)² + 4³) => $\frac{1}{2}$(-96 + 64) => $\frac{1}{2}$(-32) => -16, i.e. Area is 16

Ans.: (ii) x = 2, Area = 8

 

[SithsNGiggles]

Let the point $b(x_0, y_0) = (x, 0)$.

$A = \frac{1}{2} (base)(height)$
$A = \frac{1}{2} (6 - x)(x^2 - 0)$
$A = \frac{1}{2} (6x^2 - x^3)$
$A = 3x^2 - \frac{1}{2}x^3$

$A' = 6x - \frac{3}{2} x^2$
$A' = \frac{3}{2} x (4 - x)$

Setting $A'= 0$,
$\frac{3}{2} x (4 - x) = 0$
$x = 0, x = 4$

Applying the first derivative test:
On the interval $(0, 4), A' > 0$, which indicates that $A$ is increasing on this interval.
On $(4, 6), A' < 0$, so $A$ is decreasing.
The increase-decrease indicates the occurrence of a maximum at $x = 4$.

And when $x = 4, y = 4^2 = 16$. The area is then
$A = \frac{1}{2} (6 - 4) (16) = 16$.

If I haven't made any mistakes here, my best guess would be that it's a typo in your book.