document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1083631>Question 1083631</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The cross section of a trough is an isosceles trapezoid. If the trough is made by bending up the sides of a strip of metal C inches wide, what should be the angle of inclination of the sides and the width across the bottom if the cross-sectional area is to be a maximum?\r<BR>\n" );
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document.write( "Answer: Angle= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+pi%2F3+ ALIGN=MIDDLE  ALT=\"+pi%2F3+\"   DISABLED_style= \"max-width:90%; height:auto;\" >;<BR>\n" );
document.write( "width across the bottom is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+c%2F3+ ALIGN=MIDDLE  ALT=\"+c%2F3+\"   DISABLED_style= \"max-width:90%; height:auto;\" > ; <BR>\n" );
document.write( "maximum area is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+c%5E2%2F%284%2Asqrt%28+3+%29%29+ ALIGN=MIDDLE  ALT=\"+c%5E2%2F%284%2Asqrt%28+3+%29%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Thank you, I promise work in some problems coming soon. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #698049 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=698049\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/display-illustration.mpl?tutor=htmentor&illustration_name=trapezoid3\"><BR>\n" );
document.write( "The area of the trapezoid pictured above is H(W+X)<BR>\n" );
document.write( "We need to express this in terms of the angle a and the width W<BR>\n" );
document.write( "From the figure we see that H = Lcos(a) and X = Lsin(a)<BR>\n" );
document.write( "Since C = W + 2L we have L = (C-W)/2<BR>\n" );
document.write( "Thus H = 1/2(C-W)cos(a) and X = 1/2(C-W)sin(a)<BR>\n" );
document.write( "Hence the area A = (W/2)(C-W)*cos(a)+((1/2)(C-W))^2*sin(a)*cos(a)<BR>\n" );
document.write( "To maximize the area in terms of both W and a, we take the partial derivatives <BR>\n" );
document.write( "w.r.t. each variable and set equal to zero.<BR>\n" );
document.write( "First we maximize in terms of W:<BR>\n" );
document.write( "dA/dW = 0 = (C/2 - W)*cos(a) + (1/4)sin(a)*cos(a)(2W - 2C)<BR>\n" );
document.write( "(C/2 - W)*cos(a) - (1/2)(C - W)sin(a)*cos(a) = 0<BR>\n" );
document.write( "Multiply through by 2 and divide through by cos(a):<BR>\n" );
document.write( "C - 2W - (C - W)sin(a) = 0<BR>\n" );
document.write( "Solve for sin(a):<BR>\n" );
document.write( "sin(a) = (C - 2W)/(C - W) [1]<BR>\n" );
document.write( "Now maximize the area in terms of a:<BR>\n" );
document.write( "dA/da = 0 = (W/2)(C-W)(-sin(a)) + (1/2)(C-W)^2(cos^2(a)-sin^2(a))<BR>\n" );
document.write( "Using the identity cos^2(a) = 1 - sin^2(a), we can write<BR>\n" );
document.write( "(-W/2)(C-W)sin(a) + (1/2)(C-W)^2(1 - 2sin^2(a)) = 0<BR>\n" );
document.write( "Divide through by (C-W)/2:<BR>\n" );
document.write( "-W*sin(a) + (C-W)(1 - 2sin^2(a)) = 0<BR>\n" );
document.write( "Substitute the value for sin(a) in [1] above:<BR>\n" );
document.write( "-W*(2W-C)/(W-C) + (C-W)(1-2((2W-C)/(W-C))^2)<BR>\n" );
document.write( "Carrying out the algebra and collecting like terms we are left with:<BR>\n" );
document.write( "2W - C - (W-C)/2 = 0<BR>\n" );
document.write( "2W - W/2 - C/2 -> W = C/3<BR>\n" );
document.write( "Use this expression to determine the value of the angle a:<BR>\n" );
document.write( "sin(a) = (2(C/3) - C)/(C/3 - C) = 1/2 -> a = pi/6<BR>\n" );
document.write( "This means that the inclination angle from the horizontal is pi/2 - pi/6 = pi/3<BR>\n" );
document.write( "Substituting the values for sin(a), cos(a) and W, we can solve for the area in terms of C:<BR>\n" );
document.write( "A = (C/3)((C-(C/3))/2)*sqrt(3)/2 + (1/2)((C-C/3)/2)^2*sqrt(3)/2<BR>\n" );
document.write( "This gives A = C^2/(4*sqrt(3))\r<BR>\n" );
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