#15: OPTIMIZATION

Optimization is the process of finding values that make a given quantity the largest - or smallest - possible within certain constraints. If you can write the given quantity as a function of a single variable, then you can use calculus to find the maximum or minimum value.

1. Situation: You want to make an ornate mirror. You have decided that the shape of the frame will be elliptical, two feet wide and 3 feet tall. The mirror portion will be a rectangle inscribed inside the ellipse. You want to maximize the area of the mirror. That is, you want to find the dimensions of the rectangle of greatest area which can be inscribed in the frame.

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  2. You want to maximize the area of the rectangular section. Since the mirror is symmetric, you can look at the upper right quarter of the mirror, and maximize its area. Call the quarter area A
  3. The size and shape of the frame is fixed. If you put the origin of your coordinate system in the center of the mirror, then find the equation of the edge of the frame.
  4. The corner of the rectangle must be on the edge of the frame. From the equation you have just written, solve for one of the variables in terms of the other.
  5. The area of the quarter mirror is A = xy. Use the relationship you found between x and y to find the area in terms of one variable.
  6. Determine what values of your variable make sense for this problem.
  7. Find the critical points of A.
  8. Make a table to determine the maximum area.
  9. Find the dimensions of the rectangular portion of the mirror.



2. Situation: You are doing some duct work in a building. You need to get a long piece of metal to the area you are working in. To get to this area you need to crawl through the duct which has a right angle turn in it. The width of the duct before the turn is 3 feet. The width after the turn is 2.5 feet. What is the longest piece of metal you can get around the corner? (Assume the width of the metal can fit into the duct and the metal can't bend)

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  2. You want to find the length of the longest piece of metal that can fit around the corner.
  3. The fixed variables are the dimensions of the duct and that the metal must be straight (can't bend). To determine equations to formulate this problem, think about what happens to the piece of metal if it is too big. The worst case is when the metal just makes it around the corner, touching both walls and the inner corner as it squeezes by. Therefore, if you find the minimum distance of a line touching both walls and the inner corner, that will give you the biggest piece of metal that will fit around the corner. With this new approach, you are looking at the length of a line segment between the intercepts touching both walls and passing through the inner corner which is the point (2.5,3). Draw a picture representing this.
  4. The slope of the line, m, will vary as well as where the line touches the walls, at x = a and y = b. Write the equation of the line through the point (2.5,3) in terms of the slope m.
  5. Find the intercepts, a and b, in terms of m.
  6. Find the formula for the length of the line s in terms of a and b.
  7. Write s in terms of b.
  8. Find the domain of s.
  9. Find the critical points for s.
  10. Make a table.
  11. Find the length of the largest piece of metal that can get around the corner.



TRY IT #1: Rework all the above steps in the previous problem except find the length of the line s in terms of a. Verify that you get the same length for the piece of metal.

  1. Write the intercept form of a straight line.
  2. Use the fact that the line goes through the point (2.5,3) to find b in terms of a.
  3. Find the formula for the length of the line s in terms of a and b.
  4. Write s in terms of a.
  5. Find the domain of s.
  6. Find the critical points for s.
  7. Make a table.
  8. Find the length of the largest piece of metal that can get around the corner.



rbyrne@math.cudenver.edu

ROXANNE BYRNE :UNIVERSITY OF COLORADO AT DENVER: ©:2002, Roxanne Byrne