SOLUTION: A culvert designed with a semicircular cross-section of diameter d is to be
redesigned to have an isosceles trapezoidal cross-section (angled sides are
of equal length) by inscri
Algebra.Com
Question 972903: A culvert designed with a semicircular cross-section of diameter d is to be
redesigned to have an isosceles trapezoidal cross-section (angled sides are
of equal length) by inscribing the trapezoid in the semi circle.
What is the length of the bottom base x of the trapezoid if its area is to be
a maximum? What is the maximum area of the trapezoidal cross-section in
terms of d?
Hint: First find an expression for the area of cross section in terms of x
and d.
Amy.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
First draw the picture. Bottom half of a circle with center O. Sketch in the trapezoid. Points of intersection of the bottom base of the trapezoid and the semi-circle are A and B. Drop a vertical radius from O that bisects the bottom base. Point of intersection of this radius and the bottom base is C. Construct radius OA.
Note the right triangle OAC. The hypotenuse is the radius OA which measures 
. Since OC bisects AB, the short leg of the triangle measures 
. Then Mr. Pythagoras assures us that the other leg of the triangle, which is the height of the trapezoid, measures 
.
Since the area of a trapezoid is the average of the two bases times the height, we can hold 
constant and write the function for the area of the trapezoid in terms of 
thus:
\ =\ \left(\frac{d\ +\ x}{2}\right)\left(\frac{\sqrt{d^2\ -\ x^2}}{2}\right))
Taking the first derivative (use the product rule and a bunch of ugly algebra that I will leave as an exercise for the student)
\ =\ \frac{2x^2\ +\ dx\ +\ d^2}{4\sqrt{d^2\ -\ x^2}})
Set the first derivative equal to zero and solve for in terms of , we get (after multiplying through by the denominator)
Then a little quadratic formula work the details of which you can work out, and then discarding the absurd negative root, we end up with:
Then plug this value for back into the area function that we previously derived and simplify. Again, I leave the details to you, but you should end up with:
Or not quite 32 and a half square feet if you started with a 10 foot diameter culvert.
Piece of cake; easy as pie. Or is that piece of pie, easy as cake...I can never remember.
John
My calculator said it, I believe it, that settles it
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