SOLUTION: diff calc:A trapezoidal gutter is to be made from a strip of metal 22 m wide by bending up the sides. If the base is 14 m, what width across the top gives the greatest carrying cap
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Question 1048507: diff calc:A trapezoidal gutter is to be made from a strip of metal 22 m wide by bending up the sides. If the base is 14 m, what width across the top gives the greatest carrying capacity?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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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 
. (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 
. The average of the bases is \ +\ 14}{2}\ =\ 14\ +\ x)
. Hence, the area of the trapezoid as a function of the measurement 
is:
\ =\ \left(14\ +\ x\right)\sqrt{16\ -\ x^2})
Use the Product Rule:
}{\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 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:
^{\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
My calculator said it, I believe it, that settles it
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