Question 1161281: A person needed a new rain spout, but instead of buying one, she
decided to make her own by bending up an equal amount on both sides
of a piece of sheet metal. If the sheet metal was 20 inches long,
how much would she have to bend up on both sides so that the area
would be 50 square inches? If you have two answers, type in the larger
of your answers.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! What area? The bottom area? The cross section?
Is the sheet of metal square? Rectangular in shape?
A good design is like this,  with the bends marked as red lines, at  inches from the edges.
Bending up the sides so they are perpendicular to the middle section,
you would get a channel to carry the water.
(Considering other bending angles could not be the intention of the problem, as it would complicate it and lead to an infinite number of solutions)
The cross-section would look like this: 
The area of the rectangular cross-section is a function of ,
  , or  .
The greater that area, the more water that rain spout can carry per minute.
For a cross-section area of 50 square inches, should satisfy
 , and that is the quadratic equation to solve.
(The hint about "two answers" suggested a quadratic equation).
If you have been shown a way to solve a problem like that in class,
just do it as your teacher suggests.
If that involves transforming that equation into a preferred form,
and/or applying a memorized formula, pleasing the teacher is advisable.
Solving it by factoring or by "completing the square' may be also accepted, and it is easy enough in this case.
If you are left to your own resources, you could try different values of aiming to get  .
Those values must be between  and  , because the lengths and  must be positive).
The function  is a quadratic function with a nice symmetrical graph called a parabola.
The function has a maximum (represented by the top of that symmetrical graph.
The value  could be reached at the maximum (one solution),
or at two points to either side of that maximum (2 solutions),
or never (if the maximum value is less than 50).
You realize that  ,
so its value is zero at  and 
(obvious if you thinks what that means for the cross-section drawing).
Halfway between those two symmetrical points on the graph is the maximum,
with  and  .
So, the get a cross-section area of  ,
you need to bend up  on both sides,
and is the maximum cross section area possible.
|
|
|
