SOLUTION: 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.
a. Express
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a. Express
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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.
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)
B. For what value of x is the volume of the gutter a maximum? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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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It will help to draw this out. Label the height of the gutter h, sides of the gutter as x
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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)
The area of a trapezoid: A =
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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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x\__/x, is a picture of the end of the gutter
We have to find a, b and h in terms of x
b = (60-2x), the shortest width of the gutter
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a = the longer width of the gutter: this width is the shorter width plus the sides of the right triangle formed on both sides.
the angle of this right triangle: 120-90 = 30 degrees.
Find the length of this side (s) in terms of x
sin(30) = , the side opposite
s = .5x
a = (60-2x) + 2(s)
replace s with .5x
a = 60 - 2x + 2(.5x)
a = 60 - 2x + x
a = (60 - x) is the longer side of the trapezoid
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h = the height of the gutter is the adjacent side of this same right triangle
cos(30) =
h = .866x
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Substituting in A =
A =
A = .433x(120 - 3x)
A = 52x - 1.3x^2 sq cm is the area of the end of the gutter
To find the volume multiply this by 1000 cm (10 meters)
V = 1000(52x - 1.3x^2)
V = 52000x - 1300x^2 cu/cm is volume of the gutter
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B. For what value of x is the volume of the gutter a maximum?
The above is a quadratic equation.
the max occurs at the axis of symmetry which is x = -b/(2a)
x =
x = 20 cm is the sides on the gutter for max volume
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Let me know if you get this OK. ankor@att.net
