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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 $4y = 16 - x^2$
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closed as off-topic by Saad, user284331, The Chaz 2.0, Cave Johnson, JonMark Perry Apr 6 '18 at 2:57

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  • $\begingroup$ 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? $\endgroup$ – Doug M Apr 5 '18 at 21:25
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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 $x$

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  • $\begingroup$ Nice answer. It will be fine if you explain also that why you consider an isosceles trapezoid. $\endgroup$ – Qurultay Apr 5 '18 at 21:29
  • $\begingroup$ 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... $\endgroup$ – Daniel Gendin Apr 5 '18 at 21:31
  • $\begingroup$ @DanielGendin thanks, fixed $\endgroup$ – Doug M Apr 5 '18 at 21:40
  • $\begingroup$ @Qurultay It is given that one base is the x, axis, and the parabola is symmetric about the y axis, the parabola must be isosceles. $\endgroup$ – Doug M Apr 5 '18 at 21:42

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