Largest-area trapezoid bounded between x-axis and the quadratic function $4y = 16 - x^2$? [closed]

Calculus optimization problem:

Find the dimensions of the largest possible trapezoid, by area, that fulfills the following criteria:

• longer base runs along the x-axis
• other two vertices sit above the x-axis
• bounded by the quadratic function $4y = 16 - x^2$

closed as off-topic by Saad, user284331, The Chaz 2.0, Cave Johnson, JonMark PerryApr 6 '18 at 2:57

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• Any ideas? What is the range of values that the vertices might fall in? Can you find equations for the vertices? The area of the a trapezoid inscribed between these curves. And once you have a formula for area, how do you maximize it? – Doug M Apr 5 '18 at 21:25

For any $x \in (-4,4)$ the vertices are $(-4,0), (-x,\frac {16-x^2}{4}),(x,\frac {16-x^2}{4}),(4,0)$
$A = \frac 12 (8+2x)(\frac {16-x^2}{4})\\ \frac {dA}{dx} = ??? = 0$
Solve for $x$
• I think you are missing some devisions by 4 since the parabola is $4y=16-x^2$. It doesen't matter as it will lead to the same answer but still... – Daniel Gendin Apr 5 '18 at 21:31