document.write( "Question 995547: a trapezoidal gutter is to be made, from a strip of metal 12m wide by bending up the edges. Determine where to bend the metal and the angle to bend it at to maximise the cross sectional area and hence the capacity of the gutter
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Algebra.Com's Answer #614384 by KMST(5328) $\"\"$ $\"About$

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The fact that the problem asks about \"the angle\" to bend the metal,
\n" ); document.write( "means that the trapezoid section is an isosceles trapezoid,
\n" ); document.write( "with the same angle on either side of each base.
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with

\n" ); document.write( "The area is a function of $\"x\"$ and $\"y\"$ .
\n" ); document.write( "Because of its definition as $\"cos%28A%29\"$ , $\"-1%3C=x%3C=1\"$ ,
\n" ); document.write( "but $\"x=1\"$ would give us a zero area,
\n" ); document.write( "and we know that for $\"x%3C0\"$ , $\"-x\"$ would give us a larger area,
\n" ); document.write( "so we expect that for maximum area we need $\"0%3C=x%3C1\"$ .
\n" ); document.write( "Also, we know that $\"0%3C=y%3C=6\"$ ,
\n" ); document.write( "and since $\"y=0\"$ would give us a zero area,
\n" ); document.write( "we expect that for maximum area we need $\"0%3Cy%3C=6\"$ .
\n" ); document.write( "We need to find a maximum for the function
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\n" ); document.write( "in the domain $\"system%280%3C=x%3C1%2C0%3Cy%3C=6%29\"$ .
\n" ); document.write( "Towards the boundaries $\"x=1\"$ and $\"y=0\"$ , $\"area\"$ tends to zero.
\n" ); document.write( "On each of the boundaries $\"x=0\"$ and $\"y=6\"$ ,
\n" ); document.write( " $\"area\"$ is a function of the other variable,
\n" ); document.write( "with a maximum of $\"18\"$ .
\n" ); document.write( "Is there a local maximum, with $\"area%3E18\"$ somewhere in the middle of the domain?
\n" ); document.write( "If there is one, the partial derivatives at that point are zero.
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\n" ); document.write( " $\"d%28area%29%2Fdy=%2812-4y%2B2xy%29sqrt%281-x%5E2%29%29=2%286-2y%2Bxy%29sqrt%281-x%5E2%29%29\"$
\n" ); document.write( " $\"system%28d%28area%29%2Fdx=0%2C0%3C=x%3C1%2C0%3Cy%3C=6%29\"$ ---> $\"y-12x%2B2xy-2x%5E2y=0\"$
\n" ); document.write( " $\"system%28d%28area%29%2Fdy=0%2C0%3C=x%3C1%2C0%3Cy%3C=6%29\"$ ---> $\"6-2y%2Bxy=0\"$
\n" ); document.write( "So we have to solve
\n" ); document.write( " $\"system%286-2y%2Bxy=0%2Cy-12x%2B2xy-2x%5E2y=0%29\"$
\n" ); document.write( " $\"6-2y%2Bxy=0\"$ ---> $\"6=2y-xy\"$ ---> $\"6=%282-x%29y\"$ ---> $\"y=6%2F%282-x%29\"$
\n" ); document.write( "Substituting $\"y=6%2F%282-x%29\"$ 0 into $\"y-12x%2B2xy-2x%5E2y=0\"$ we get
\n" ); document.write( " $\"6%2F%282-x%29-12x%2B12x%2F%282-x%29-12x%5E2%2F%282-x%29=0\"$
\n" ); document.write( " $\"6%2F%282-x%29-12x%282-x%29%2F%282-x%29%2B12x%2F%282-x%29-12x%5E2%2F%282-x%29=0\"$
\n" ); document.write( " $\"6%2F%282-x%29%2B%2812x%5E2-24x%29%2F%282-x%29%2B12x%2F%282-x%29-12x%5E2%2F%282-x%29=0\"$
\n" ); document.write( " $\"%286%2B12x%5E2-24x%2B12x-12x%5E2%29%2F%282-x%29=0\"$
\n" ); document.write( " $\"%286-12x%29%2F%282-x%29=0\"$ ---> $\"6%281-2x%29%2F%282-x%29=0\"$ ---> $\"1-2x=0\"$ ---> $\"highlight%28x=1%2F2%29\"$ ---> $\"cos%28A%29=1%2F2\"$ ---> $\"highlight%28A=pi%2F3%29\"$ or $\"highlight%28A=60%5Eo%29\"$ .
\n" ); document.write( "Then $\"system%28x=1%2F2%2Cy=6%2F%282-x%29%29\"$ ---> $\"highlight%28y=4%29\"$ .
\n" ); document.write( "So bending lengthwise the 12m wide strip of metal at $\"highlight%2860%5Eo%29\"$ angles along two lines $\"highlight%284m%29\"$ from the edges, you get a gutter with a section in the shape of an isosceles trapezoid with maximum area.
\n" ); document.write( "That maximum area is
\n" ); document.write( " $\"area=6%2A2sqrt%283%2F2%29m%5E2=13sqrt%283%29=about20.8m%5E2\"$

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