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Homework Help: Related Rates & Optimisation (challenging)

  1. Apr 25, 2007 #1
    1. The problem statement, all variables and given/known data
    The first problem is a related rates question:

    A swimming pool if 5m wide and 25m long, 1m deep at the shallow end and 4.5 deep at the deepest point. A cross section is shown in the figure below. The pool is being filled at a rate 0.5m^3/hr. Use calculus techniques to find out how fast is the water level rising when the depth at the deepest point is 2.5m.

    I understand we want to find dh/dt when h at the deepest point is 2.5m. I know we are provided with dv/dt but I am absolutely clueless as to how to find out dh/dv!

    http://img441.imageshack.us/img441/7691/untitledgg5.jpg [Broken]

    The second question I found even more challenging...

    A closed box of a square base and rectangular sides is to have a volume of 1000cm^3. If the material used for the dies of the box is 20% more expensive per square meter than the material on the bottom, and the material to produce the top costs 50% more per square meter than that of the bottom, find the most economical proportions of the box.

    I'm totally clueless in this one and have absolutely no idea where to start. :confused: :bugeye: :eek:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 25, 2007 #2
    There's too much geometry in the first one for me at this time of the night.

    For the second one, there are 2 variables - the length and the height of the box. The volume has to stay constant and hopefully you can express the volume with your 2 variables. That gives you your first equation.

    Your second equation is found from cost considerations. Add up the costs for the bottom, sides and top of the box while taking into account the material cost. Do this by calculating the area of each portion, and multiplying it by its corresponding cost/unit area (1 for bottom, 1.2 for sides, 1.5 for top).

    You now have your second equation, eliminate a variable from it with your first equation. Now use calculus to minimize the cost.
    Last edited: Apr 25, 2007
  4. Apr 25, 2007 #3
    I'm a bit confused, but would that mean the side areas would be 1.2xy, and the top area would be 1.5x^2?

    Meaning total surface area (x^2) + 4(1.2xy) + (1.5x^2) ?
  5. Apr 25, 2007 #4


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    Look at your picture. When the depth is 2.5 m, the water is part way up the sloping sides. The volume of water is the length of the pool, 5 m, times the area of the trapezoid cross section. One "base" has length 10 m. The other is the distance between the two walls at height y which depends on y. You will need to find the equation of the two lines forming the sides: x= f1(y) and x= f2(y) so the length of the top "base" is f2(y)- f1(y). Then the volume will be V= 5(1/2)y(10+f2(y)-f1(y)). You are told that dV/dt= 1/2 so you can differentiate and use that to find dy/dt.

    I would consider 2 to be much easier than the first. (I assume that "the dies of the box" is actually "the sides of the box"!).
    Take x,y,z to be the dimensions of the box where x and y are the dimensions of the bottom and z is the height of the box. Then you must have xyz= 1000. The area of the bottom is xy. Since we are only asked about "proportions", take the cost of the bottom to be 1 per unit area: the cost of the bottom is xy. The top also has has area xy but cost "50% more" than the bottom so its cost is 1.5xy. The total area of the sides is 2xz+ 2yz and the sides cost "20% more" than the bottom so the bottom cost 1.2(2xz+ 2yz). The total cost of the box is xy+ 1.5xy+ 1.2(2xz+ 2yz) and that is what you want to minimize subject to the constraint xyz= 1000. Since you showed no attempt to do this problem I don't know which of the many ways of minimizing this you have available.
    Last edited by a moderator: May 2, 2017
  6. Apr 25, 2007 #5
    Thanks for the reply. Just with the second question, when it says most economical proportions, that means we want to minimize the cost correct? That would mean I would have to manipulate xyz=1000 so that the total cost of box equation is in terms of only 1 variable?
  7. Apr 25, 2007 #6


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    Actually, because you have only one "constraint", you cannot reduce to only one variable. You can use that equation to eliminate ONE variable resulting in a function of two variables. Now, you might take PARTIAL derivatives with respect to the two remaining variables and set them equal to 0 or just use the "Lagrange multiplier" method on the original 3 variable method. Yet another way is to argue "by symmetry" that two of the variables, the dimensions of the top and bottom, must be the same so that you have only 2 distinct variables, then use xyz= x2z= 1000 to reduce to one variable.
  8. Apr 25, 2007 #7
    I understood none of that except the converting xy to x^2 part:rofl:

    does that also mean that in the total cost equation I can convert all x's to y's or vice versa?

    If so, what was the point of letting one side of the square = x and the other side of the square = y?
    Last edited: Apr 25, 2007
  9. Apr 25, 2007 #8


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    Yes, "most economical" means "least cost".

    Since you only have one "constraint" you can only remove one variable- you can't immediately reduce to only one variable. As I said before, I don't know what techniques you have available.

    You might use the "Lagrange multiplier" method on the original three variable problem, or

    use xyz= 1000 to reduce to two variables and set the partial derivatives with respect to those two variables equal to 0, or

    argue, "by symmetry", that the dimensions of the top and bottom, x and y, must be equal, then use xyz= x2z= 1000 to eliminate z, leaving only x as the variable. That last is simplest, though you might have some difficulty convincing your teacher that the symmetry argument is valid.
  10. Apr 26, 2007 #9

    I'm not quite following. :confused: The base is a square meaning so is the top! That's what I've done.

    I've let x²z=1000 and then transposed to get z=1000/x². Then substituted z=1000/x² into the total cost equation that was T=x² + 1.2x² + 1.5xy. Then differentiated and solved to find the value for x for which cost is minimized. Finally I subbed that minmized x value back into z=1000/x² to find the value for z for which cost is minimized.

    Here to hoping it's that easy..:surprised
  11. Apr 26, 2007 #10
    Sorry to be a hassle but I'm also struggling to comprehend how you find the other base! :confused: :cry:
    Last edited: Apr 26, 2007
  12. Apr 26, 2007 #11
    Bump! Anybody out there?:cry:
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