Question 565891
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Before I even start, I want you to go back and read the question as you posted it.  "...shown in the illustration."  Right.  This is Algebra.com -- NOT the Psychic Hot Line.  First lesson:  Mathematics requires using your head for something besides a hat rack.


Be that as it may, I'm going to presume that you want to fold up two sides making an unknown depth (equal on both sides) on a trough with no top and a bottom that perforce measures 24 inches minus 2 times the depth.  To the extent that my assumptions about the true nature of your problem are correct, begin with the concept that the area of a rectangle is given by length (or depth in this case) times width. 


For this problem we let *[tex \Large x] represent the amount folded up or the depth of the trough, and then the width of the rectangle (the bottom of the trough must be *[tex \Large 24\ -\ 2x].  Multiplying length times width gives area as a function of the depth:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ 24x\ -\ 2x^2]


Written in standard *[tex \Large \left(\rho(x)\ =\ ax^2\ +\ bx\ + c\right)] form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ -2x^2\ +\ 24x]


<i><b>Algebra Solution</b></i>  Recognize that this is a quadratic polynomial function that will graph to a parabola.  Since the lead coefficient is negative, the parabola must open downward, hence the vertex is a maximum.  Using *[tex \Large x_v\ =\ \frac{-b}{2a}\ =\ \frac{-24}{2(-2)}\ =\ 6].  Hence, fold up 6 inches to get the maximum area, giving you a 6 by 12 (24 - 12) rectangle.


<i><b>Calculus Solution</b></i>  Take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dx}\ =\ -4x\ +\ 24]


Set the first derivative equal to zero and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4x\ +\ 24\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 6]


Hence extreme point at *[tex \Large x\ =\ 6]


Take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dx^2}\ =\ -4\ <\ 0\ \forall\ x] in the domain of A.


Hence the extreme point is a maximum.


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