Application of Extrema: metal sheet folded into trough

[yoyosuper7]

agh! I have the Application of Extrema problem and I have no idea how to solve. Can someone please help me?

A sheet of aluminum 30 inches wide is to be used to make a trapezoidal trough by turning up the edges of the sheet. The two sides formed will be the same length (x), leaving the bottom of the trough 30 - 2x inches wide. The sides make an angle [theta] with the ground outside the trough. Find x and [theta] to maximize the cross section (and hence the volume) of the trough. Find the corresponding maximum cross sectional area. If the goal is to maximize cross section, would it be better to bend the aluminum into a semicircular cross-section than a trapezoidal one?

Thanks!!!

John

 

[soroban]

Hello, John!

I solved this at another site, but for those just tuning in . . .

A sheet of aluminum 30 inches wide is to be used to make a trapezoidal trough
by turning up the edges of the sheet.
The two sides formed will be the same length \(\displaystyle x\),
leaving the bottom of the trough \(\displaystyle 30 - 2x\) inches wide.
The sides make an angle \(\displaystyle \theta\) with the ground outside the trough.
Find \(\displaystyle x\) and \(\displaystyle \theta\) to maximize the cross section (and hence the volume) of the trough.
Find the corresponding maximum cross sectional area.

If the goal is to maximize cross section, would it be better to bend the aluminum
into a semicircular cross-section than a trapezoidal one? . . . . Yes!

    : x cos @ :  30 - 2x  : x cos @ :
    * - - - - * - - - - - * - - - - *
    :\        :           :        /:
    : \       :           :       / :
    :  \      :           :      /  :
    :   \ x   :           :   x /   :
    :    \    :           :    /    :
    :     \   :           :   /     : x sin @
    :      \  :           :  /      :
    :       \ :           : /       :
    :        \:           :/ @      :
    * - - - - * - - - - - * - - - - * 
    : x cos @ :  30 - 2x  : x cos @ :

\(\displaystyle \text{The area of a trapezoid is: }\:A \:=\:\frac{h}{2}\left(b_1 + b_2\right)\)

We have: \(\displaystyle h \:=\: x\sin\theta\)
. . . . . . . \(\displaystyle b_1 \:=\: 30-2x\)
. . . . . . . \(\displaystyle b_2 \:=\: 30-2x + 2x\cos\theta\)

\(\displaystyle \text{Then: }\;A \;=\;\tfrac{1}{2}(x\sin\theta)\bigg[(30-2x) + (30-2x + 2x\cos\theta)\bigg]\)

. . \(\displaystyle \text{which simplifies to: }\;A \;=\;30x\sin\theta - 2x^2\sin\theta + x^2\sin\theta\cos\theta\)


Equate partial derivatives to zero, and solve . . .

. . \(\displaystyle \dfrac{\partial A}{\partial x} \;=\;30\sin\theta - 4x\sin\theta + 2 x\sin\theta\cos\theta \;=\;0\)

. . \(\displaystyle \dfrac{\partial A}{\partial\theta} \;=\;30x\cos\theta - 2x^2\cos\theta + x^2(\cos^2\!\theta - \sin^2\!\theta) \;=\;0\)


\(\displaystyle \text{It took a while, but I got: }\:x \:=\: 10,\;\theta \:=\:60^o\)

 

[galactus]

That's what I got as well, Soroban. The max is attained when the sides are equal. I deleted my post because I did not feel it was thorough enough.

This one is a little tougher than most because no side length is given.

 

[yoyosuper7]

Ohh! thanks guys, I completly understand the problem now. I've always had a hardtime solving word problems, but thanks to you, everything is a lot clearer now and makes sense. Thanks a lot!!!