document.write( "Question 1011898: A sheet of metal is 60 cm wide and 10 m long. It is bent along its width to form a gutter with a cross section that is an isosceles trapezoid with 120 degree angles. \r
\n" ); document.write( "\n" ); document.write( "a. Express the volume V of the gutter as a function of x, the length in cm of one of the equal sides. ( hint: Volume= area of trapezoid x length of gutter)\r
\n" ); document.write( "\n" ); document.write( "B. For what value of x is the volume of the gutter a maximum?
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Algebra.Com's Answer #627772 by ankor@dixie-net.com(22740)\"\" \"About 
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A sheet of metal is 60 cm wide and 10 m long. It is bent along its width to form a gutter with a cross section that is an isosceles trapezoid with 120 degree angles.
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\n" ); document.write( "It will help to draw this out. Label the height of the gutter h, sides of the gutter as x
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\n" ); document.write( "a. Express the volume V of the gutter as a function of x, the length in cm of one of the equal sides. ( hint: Volume= area of trapezoid x length of gutter)
\n" ); document.write( "The area of a trapezoid: A = \"%28h%2F2%29%28a%2Bb%29\"
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\n" ); document.write( "Ahh yes, sorry! Can I blame it on New year's eve wine? No? Ok let me try to straighten this out.
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\n" ); document.write( " x\__/x, is a picture of the end of the gutter
\n" ); document.write( "We have to find a, b and h in terms of x
\n" ); document.write( "b = (60-2x), the shortest width of the gutter
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\n" ); document.write( "a = the longer width of the gutter: this width is the shorter width plus the sides of the right triangle formed on both sides.
\n" ); document.write( " the angle of this right triangle: 120-90 = 30 degrees.
\n" ); document.write( " Find the length of this side (s) in terms of x
\n" ); document.write( "sin(30) = \"s%2Fx\", the side opposite
\n" ); document.write( "s = .5x
\n" ); document.write( "a = (60-2x) + 2(s)
\n" ); document.write( "replace s with .5x
\n" ); document.write( "a = 60 - 2x + 2(.5x)
\n" ); document.write( "a = 60 - 2x + x
\n" ); document.write( "a = (60 - x) is the longer side of the trapezoid
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\n" ); document.write( "h = the height of the gutter is the adjacent side of this same right triangle
\n" ); document.write( "cos(30) = \"h%2Fx\"
\n" ); document.write( "h = .866x
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\n" ); document.write( "Substituting in A = \"%28h%2F2%29%28a%2Bb%29\"
\n" ); document.write( "A = \"%28.866x%2F2%29%2860-x%29+%2B+%2860-2x%29%29\"
\n" ); document.write( "A = .433x(120 - 3x)
\n" ); document.write( "A = 52x - 1.3x^2 sq cm is the area of the end of the gutter
\n" ); document.write( "To find the volume multiply this by 1000 cm (10 meters)
\n" ); document.write( "V = 1000(52x - 1.3x^2)
\n" ); document.write( "V = 52000x - 1300x^2 cu/cm is volume of the gutter
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\n" ); document.write( "B. For what value of x is the volume of the gutter a maximum?
\n" ); document.write( "The above is a quadratic equation.
\n" ); document.write( "the max occurs at the axis of symmetry which is x = -b/(2a)
\n" ); document.write( "x = \"%28-52000%29%2F%282%2A-1300%29\"
\n" ); document.write( "x = 20 cm is the sides on the gutter for max volume
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\n" ); document.write( "Let me know if you get this OK. ankor@att.net
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