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Math Help - Functions of Several Variables

  1. #1
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    Functions of Several Variables

    Hello all,

    I dont think my lecturer illustrate the application of functions of several variables generously. I dont understand how to apply it, such as in the question below: ( I believe it is actually an easy question)

    Its winter season. Let T(x,y)= 36 - (1/5)[x^2 + (y - 5)^2] be the temperature at location(x,y) in a 10ft x 10ft hotel room with a heater on at night. One corner of the room is at (0,0) and the opposite corner is at (10,10).

    i)What is the domain of the temperature function?

    ii)Where is the likely location of the heater?

    iii)Suppose you like to sleep within the temperature range of 20celcious to 25celcious. Where would you put your bed?

    iv)Determine the locations in the room where the temperature is lowest.

    An advanced thank you for all the help. Really appreciate it. Ironically I manage to do everything else on the tutorial's questions except this first question.Thx!
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  2. #2
    MHF Contributor
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    Here is one way.

    I don't know how your lecturer explained the topic and how to solve problems regarding them. He might be using partial derivatives or whatever.

    This way here is like treating the multi-variable function the way I approach single-variable functions. It might be different from your lecturer's approach.

    T(x,y)= 36 - (1/5)[x^2 + (y - 5)^2]

    i)What is the domain of the temperature function?
    The domain is what defines the function.
    What values of x and y in the 10 by 10 room that will give a real T(x,y)?
    x = 0 to 10, and y = 0 to 10 also
    So, domain is all values of x, and/or y, from 0 to 10 ft.
    x = [0,10] ft.
    y = [0,10] ft.

    ii)Where is the likely location of the heater?
    T(x,y) has a [x^2 +(y-5)^2] in it.
    When that is equal to zero, T is maximum, T = 36 degrees Celsius.
    To make that zero, then x=0 and y=5.
    I say the likely location is at (0,5).

    iii)Suppose you like to sleep within the temperature range of 20celcious to 25celcious. Where would you put your bed?
    I'd solve for area in the room where the temperature is between 20 and 25 degrees Celsius.

    When T = 20 degree:
    20 = 36 -(1/5)[x^2 +(y-5)^2]
    -16 = -(1/5)[x^2 +(y-5)^2]
    80 = x^2 +(y-5)^2 ---------------(1)
    That is a circle whose center is at (0,5), and whose radius is sqrt(80) = 8.94 ft.

    When T = 25 degree:
    25 = 36 -(1/5)[x^2 +(y-5)^2]
    -11 = -(1/5)[x^2 +(y-5)^2]
    55 = x^2 +(y-5)^2 ---------------(1)
    That is a circle whose center is at (0,5), and whose radius is sqrt(55) = 7.42 ft.

    Therefore, I will place my bed within the area that is 7.42 ft and 8.94 ft away from the heater. The area's inner radius is 7.42 ft; the outer radius is 8.94 ft.

    iv)Determine the locations in the room where the temperature is lowest.
    They are the points in the room that are farthest from (0,5).
    Just by looking, they are (10,10) and (10,0).
    The temperature there is 11 degrees Celsius.

    They are the points where [x^2 +(y-5)^2] is maximum, where T(x,y) is minimum.
    This is where partial derivatives will come in.
    Your lecturer's approach.
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  3. #3
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    Thank you so much!

    yes its partial derivative, which makes me wonder how to apply it in the question..but ur explanation has made it very clear to me..

    thx alot!!
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