Maximum Area for Irrigation ChannelDate: 02/08/2002 at 23:35:06 From: Ingrid Subject: Applied maximum and minimum problems in calculus An irrigation channel made of concrete is to have a cross section in the form of an isosceles trapezoid, three of whose sides are 4 feet long. How should it be shaped to have a maximum area? Solve as a function of x. In my class, we typically use two formulas to solve these kinds of problems, like surface area and volume. However, in this problem, I have no idea what to do after I know the area formula. Ingrid Date: 02/09/2002 at 01:56:35 From: Doctor Jeremiah Subject: Re: Applied maximum and minimum problems in calculus Hi Ingrid, Any time someone says maximum or minimum you know that calculus is involved. Look at this graph: | | | +++++ +| ^ + | | + | max + | -------------+------+------------------+-- | + | | + | | It has a maximum. But how can you tell? Well, the slope changes from positive to negative at the maximum. And to change from positive to negative the slope must pass through zero. So the maximum of the graph is where the slope equals zero. And since the slope is just the derivative, the maximum is where the derivative equals zero. So with an isosceles trapezoid we have a shape that looks like: +----------------------------------+ \ / \ / \ / 4 \ / 4 \ / \ / \ a a / +------------------+ 4 Which is the same as: 4 sin(x) 4 4 sin(x) +-------+------------------+-------+ \ | | / \ | | / \ 4 cos(x) 4 cos(x) / 4 \ | | / 4 \ | | / \x| |x/ \| |/ +------------------+ 4 Which if we rearrange the pieces we get: 4 sin(x) 4 +-------+------------------+ |\ | | |x\ | | | \ 4 cos(x) 4 cos(x) | \ | | 4 cos(x) \ | | | \x| | | \| | +-------+------------------+ 4 sin(x) 4 So the area is: A = ( 4+4sin(x) )( 4 cos(x) ) A = 4( 4 cos(x) ) + 4sin(x)( 4 cos(x) ) A = 16 cos(x) + 16 sin(x)cos(x) Notice that sin(x)cos(x) is the same as sin(2x)/2 A = 16 cos(x) + 16 sin(x)cos(x) A = 16 cos(x) + 16 sin(2x)/2 A = 16 cos(x) + 8 sin(2x) If we could just take the derivative of this, we would get the slope and we could set that to zero, find out what value of x makes the maximum area, and solve for the maximum area. d/dx( A ) = d/dx( 16 cos(x) + 8 sin(2x) ) dA/dx = d/dx( 16 cos(x) ) + d/dx( 8 sin(2x) ) dA/dx = 16 d/dx( cos(x) ) + 8 d/dx( sin(2x) ) First the first term: dA/dx = 16 d/dx( cos(x) ) + 8 d/dx( sin(2x) ) dA/dx = 16 (-sin(x)) d/dx( x ) + 8 d/dx( sin(2x) ) dA/dx = -16 sin(x) dx/dx + 8 d/dx( sin(2x) ) dA/dx = -16 sin(x) + 8 d/dx( sin(2x) ) Now the next term: dA/dx = -16 sin(x) + 8 d/dx( sin(2x) ) dA/dx = -16 sin(x) + 8 cos(2x) d/dx( 2x ) dA/dx = -16 sin(x) + 8 cos(2x) 2 d/dx( x ) dA/dx = -16 sin(x) + 16 cos(2x) d/dx( x ) dA/dx = -16 sin(x) + 16 cos(2x) dx/dx dA/dx = -16 sin(x) + 16 cos(2x) Remember, the maximum area is when the slope (derivative) is equal to zero: 0 = -16 sin(x) + 16 cos(2x) 16 sin(x) = 16 cos(2x) sin(x) = cos(2x) There is only one good answer to this: x=30 sin(x) = cos(2x) sin(30) = cos(60) So x=30 is the angle that makes the maximum area. That means the maximum area is: maxA = 16 cos(x) + 8 sin(2x) <=== x=30 maxA = 16 cos(30) + 8 sin(60) - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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