Question 1048507
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*[illustration trapezoidgutter_(2).jpg]


The carrying capacity is a function of the cross-sectional area of the gutter.  So we need to maximize the area of a trapezoid with base 1 measure of 14 and base 2 measure of *[tex \Large 14\ +\ 2x]. (See diagram).  Given 22 m of material, and the base 1 measure of 14, the measure of the sides must be 4.


Area of a trapezoid is the average of the bases times the height. So:


Pythagoras gives us the height of *[tex \Large \sqrt{16\ -\ x^2}].  The average of the bases is *[tex \Large \frac{(14\ +\ 2x)\ +\ 14}{2}\ =\ 14\ +\ x].  Hence, the area of the trapezoid as a function of the measurement *[tex \Large x] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ \left(14\ +\ x\right)\sqrt{16\ -\ x^2}]


Use the Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dx}\ =\ \sqrt{16\ -\ x^2}\ -\ \frac{x(14\ +\ x)}{\sqrt{16\ -\ x^2}}\ =\ -\frac{2x^2\ +\ 14x\ -\ 16}{\sqrt{16\ -\ x^2}]


Set the first derivative equal to zero and solve to get the critical points which represent extremum for the function.  Since *[tex \Large x] is a measure of length, discard any negative value.  The positive zero of the first derivative is the critical point of interest.  I leave this part of the algebra to you.


Use the quotient rule to take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dx^2}\ =\ \frac{2x^3\ -\ 48\ -\ 224}{\left(16\ -\ x^2\right)^{\frac{3}{2}}]


I will leave it to you to verify that the second derivative is negative for the value of the valid critical point and therefore this critical point is indeed a local maximum.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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