Optimization problem - Trapezoid under a parabola recently I've been working on a problem from a textbook about Optimization. The result that I get is $k = 8$, even thought the answer from the textbook is $k = \frac{32}{3}$
The problem follows:
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The x axis interepts the parabola $12-3x^2$ at the points $A$ and $B$, and also the line $y = k$ (for $0 < k < 12$) at the points C and D. Determine $k$ in a way that the trapezoid $ABCD$ has a maximum area.
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My solution was this
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The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{4+2 \cdot \sqrt{\frac{12-k}{3}} \cdot k}{2}$$
$$A_{T}' = 0 \therefore \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$2(12-k)-k=0 \therefore k = 8$$
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Where did I go wrong on?
Thank you guys!

The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{\left(4+2 \cdot \sqrt{\frac{12-k}{3}} \right)\cdot k}{2}$$
$$A_{T}' = 0 $$ 
$$ 2 + \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$4 \sqrt{3 \cdot (12-k)} + 24 - 3k = 0$$
$$4 \sqrt{3 \cdot (12-k)} = 3k - 24$$
Squaring both sides we get
$$16 \cdot 3 \cdot (12-k) = 9k^2-6 \cdot 24 \cdot k + 24^2 \therefore k = \frac{32}{3}$$
