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Math Help - Word Problem

  1. #1
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    Word Problem

    Imagine making a tent in the shape of a spherical cap (a sphere with a lower portion sliced away by a plane).

    Assume we want the volume to be 2.2m^3 to sleep two or three people. Draw a picture, identifying all appropriate variables. The floor of the tent is cheaper material than the rest: assume that the material making up the dome of the tent is 1.4 times as expensive per square meter than the
    material touching the ground.

    a) What should the dimensions of the tent be so that the cost of the material used is a minimum?

    b) What is the total area of the material used?

    I'm having problems simply finding the dimensions. (The radius, height, and slant) ANy advice?
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  2. #2
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    Area of spherical cap

    Hello jzellt
    Quote Originally Posted by jzellt View Post
    Imagine making a tent in the shape of a spherical cap (a sphere with a lower portion sliced away by a plane).

    Assume we want the volume to be 2.2m^3 to sleep two or three people. Draw a picture, identifying all appropriate variables. The floor of the tent is cheaper material than the rest: assume that the material making up the dome of the tent is 1.4 times as expensive per square meter than the
    material touching the ground.

    a) What should the dimensions of the tent be so that the cost of the material used is a minimum?

    b) What is the total area of the material used?

    I'm having problems simply finding the dimensions. (The radius, height, and slant) ANy advice?
    You need a formula for the surface area of a spherical cap. If the radius of the flat base is a, and the height is h, then the curved surface area is \pi(a^2 +h^2). For a diagram, and the formula see here.

    Can you do it now?

    Grandad
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  3. #3
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    Yeah...I see that. BUt that means I have one formula but two unknowns. We don't know the radius or the height. See what I mean?
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  4. #4
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    Hello jzellt
    Quote Originally Posted by jzellt View Post
    Yeah...I see that. BUt that means I have one formula but two unknowns. We don't know the radius or the height. See what I mean?
    Use the formula for the volume (on the same page) in terms of a and h. Put this equal to 2.2 and eliminate h^2.

    Grandad
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  5. #5
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    Sorry...I guess I'm giving way too much thought to a fairly simple problem.

    Okay, so using the formula and the given volume, we have:

    2.2 = 3.14h(3a^2 + h^2) / 6

    Now you say eliminate h^2? How can I just eliminate something?
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  6. #6
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    Hello jzellt
    Quote Originally Posted by jzellt View Post
    Sorry...I guess I'm giving way too much thought to a fairly simple problem.

    Okay, so using the formula and the given volume, we have:

    2.2 = 3.14h(3a^2 + h^2) / 6

    Now you say eliminate h^2? How can I just eliminate something?
    Make it the subject of one of the formulae and substitute in the other to eliminate it.

    In fact, having had a closer look at it, I think it's probably easier to eliminate a^2.

    From your equation above:

    a^2 = \frac{1}{3}\left(\frac{13.2}{\pi h}-h^2\right) (1)

    Now use the area formula (together with the area of a circle) to set up an expression for the cost of the material. If the cost of 1 unit of area of the base is 1 unit, then the cost of the dome is 1.4 units. So the total cost is

    C = \pi a^2 + 1.4\pi(a^2 + h^2)

    = \pi(2.4a^2 + 1.4h^2)

    Now substitute for a using (1), and find h for which C is a minimum (using \frac{dC}{dh}=0).

    OK now?

    Grandad
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  7. #7
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    Ok...Thank you so much. I do have one more question though...

    I don't quite understand where this equation came from:



    I get that 3.14a^2 is the area of a circle, but what is the rest and how did you come up with it?

    Thanks again for the help.
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  8. #8
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    Hello jzellt
    Quote Originally Posted by jzellt View Post
    Ok...Thank you so much. I do have one more question though...

    I don't quite understand where this equation came from:



    I get that 3.14a^2 is the area of a circle, but what is the rest and how did you come up with it?

    Thanks again for the help.
    The area of the floor is \pi a^2; the area of the dome is \pi(a^2+h^2).

    You don't actually know the cost per unit area, so (as I said before) you can assume that it's 1 unit of cost ($, , whatever) per unit area of the floor, and 1.4 units of cost per unit area of the dome. Multiply the respective areas by the cost per unit area, and add together to get the equation for C.

    OK?

    Grandad
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  9. #9
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    Ok...I'm still working with this problem. I'm on the right track but I must be missing something.

    So, after manipulating the equations to substitute in for r^2, I took the derivative, set the equation to zero and solved for h. ( I did this since I'm trying to find the MINIMUM cost)

    I got h = 2.06.

    But when I plug that h into the equation with the goal of solving for r, I end up the equation: r^2 = negative somthing.
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