Question 310349
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Because of the fact that the parabola is symmetric to the y-axis, the rectangle must also be symmetric to the y-axis.  That means that the two lower vertices are *[tex \Large \left(-x,0\right)] and *[tex \Large \left(x,0\right)].  The measure of the base of the rectangle is therefore *[tex \Large 2x].  The upper vertices, being points on the parabola are:  *[tex \Large \left(-x,16\ -\ x^2\right)] and *[tex \Large \left(x,16\ -\ x^2\right)].  Therefore, the measure of the height of the rectangle is simply *[tex \Large 16\ -\ x^2].  Therefore the area of the rectangle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ \left(2x\right)\left(16\ -\ x^2\right)\ =\ 32x\ -\ 2x^3]


Such a function has local extremes at the points where the first derivative is zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A'(x)\ =\ 32\ -\ 6x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32\ -\ 6x^2\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{\frac{32}{6}}\ =\ \pm\sqrt{\frac{16}{3}}\ =\ \pm\frac{4\sqrt{3}}{3}]


Discard the negative root since we need a positive measure of length


We are looking for a maximum, so we want to see if the value of the second derivative is negative at the extreme point.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A{''}(\frac{4\sqrt{3}}{3})\ =\ -12\left(\frac{4\sqrt{3}}{3}\right)\ <\ 0]


Which it is...


So, the horizontal dimension of the largest area rectangle is 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ \cdot\ \frac{4\sqrt{3}}{3}\ =\ \frac{8\sqrt{3}}{3}]


And the vertical dimension is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16\ -\ \left(\frac{4\sqrt{3}}{3}\right)^2\ =\ \frac{32}{3}] 



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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