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Math Help - find the minimum

  1. #1
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    find the minimum

    These are extra credit problems that i have absolutely no clue how to do. So when explaining, please use a step-by-step solution.. explaining everything. Im in pre-calc if that helps you determine what i already know.

    question #1
    For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants?

    y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

    (^ means "to the power of")

    -----------------------------------------------------------

    question #2
    Im not sure what the first symbol is but i think its an integral sign and it has
    "pi"/2 on the top of it and 0 on the bottom. directly after that it has a fraction with this on top:
    [sin(x)+3]^(1/2)

    and this is on the bottom:
    [sin(x)+2]^(1/2) + [cos(x)+3]^(1/2)

    next to all that it has "dx" and you are supposed to find what all this equals (=)
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  2. #2
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    my assignment is late now but i can still get partial credit for turning it in late. Can i please get some help? ..
    btw, i dont appreciate getting attacked and demerited for simply wanting a needed answer.. ironic, perhaps, as i fully expect to receive another point deduction on this website. A slight prick to working body is all tho.
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  3. #3
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    Quote Originally Posted by tummbler View Post
    These are extra credit problems that i have absolutely no clue how to do. So when explaining, please use a step-by-step solution.. explaining everything. Im in pre-calc if that helps you determine what i already know.

    question #1
    For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants?

    y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

    (^ means "to the power of")

    -----------------------------------------------------------

    question #2
    Im not sure what the first symbol is but i think its an integral sign and it has
    "pi"/2 on the top of it and 0 on the bottom. directly after that it has a fraction with this on top:
    [sin(x)+3]^(1/2)

    and this is on the bottom:
    [sin(x)+2]^(1/2) + [cos(x)+3]^(1/2)

    next to all that it has "dx" and you are supposed to find what all this equals (=)
    I don't think that we can give you any help that you would benefit from.
    The obvious ways to do these problems both involve calculus.

    RonL
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tummbler View Post
    These are extra credit problems that i have absolutely no clue how to do. So when explaining, please use a step-by-step solution.. explaining everything. Im in pre-calc if that helps you determine what i already know.

    question #1
    For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants?

    y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

    (^ means "to the power of")

    -----------------------------------------------------------

    question #2
    Im not sure what the first symbol is but i think its an integral sign and it has
    "pi"/2 on the top of it and 0 on the bottom. directly after that it has a fraction with this on top:
    [sin(x)+3]^(1/2)

    and this is on the bottom:
    [sin(x)+2]^(1/2) + [cos(x)+3]^(1/2)

    next to all that it has "dx" and you are supposed to find what all this equals (=)
    you're in pre-calc and doing integrals? anyway, for the integral question, i think there's a typo, shouldn't it be a sin(x) + 3 in the bottom for the first term? well, either way, its going to be complicated

    anyway, for the first question, which is kind of wierd:


    y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

    first we want to know if there are any values we can't pick. remember that square roots are not defined if what is under the square root is negative. since everything under our square roots are squares we have no problem, since any real number squared is greater than or equal to zero always.

    so now what would make it a minimum?

    consider the first square root:
    [(a^2)+(b-x)^2]^(1/2)

    a^2 is positive and constant, so we can't do anything about that, but, we can make this value as small as possible if we add 0 to a^2. since adding anything more, we get a bigger value. so we see what's added to a^2 is (b - x)^2. we want this to be zero. in order for (b - x)^2 = 0, x must be equal to b, that way we have (b - b)^2 = (0)^2 = 0.

    so for our first root, x = b yields the minimum. for the second root, the same principle applies. we want to add 0 to c^2 in order for it to be a minimum. so we must have x = 0.

    so for y to be minimum, we need x = b or x = 0. how do we know which to choose? that depends on the size of a, b, and c
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by tummbler View Post
    my assignment is late now but i can still get partial credit for turning it in late. Can i please get some help? ..
    btw, i dont appreciate getting attacked and demerited for simply wanting a needed answer.. ironic, perhaps, as i fully expect to receive another point deduction on this website. A slight prick to working body is all tho.
    And maybe we don't like being asked to do extra credit assignments for
    people who obviously should not receive that credit.

    Explaining how do do homework is one thing, we hope we are helping you
    learn. But, explaining how to do this assignment will not help you do it, as
    you have not covered the required material. All we could do is do this for you
    and that will benefit no one and undermine the maths education and
    assessment system.

    RonL
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    .. for the first question, which is kind of wierd:


    y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

    first we want to know if there are any values we can't pick. remember that square roots are not defined if what is under the square root is negative. since everything under our square roots are squares we have no problem, since any real number squared is greater than or equal to zero always.

    so now what would make it a minimum?

    consider the first square root:
    [(a^2)+(b-x)^2]^(1/2)

    a^2 is positive and constant, so we can't do anything about that, but, we can make this value as small as possible if we add 0 to a^2. since adding anything more, we get a bigger value. so we see what's added to a^2 is (b - x)^2. we want this to be zero. in order for (b - x)^2 = 0, x must be equal to b, that way we have (b - b)^2 = (0)^2 = 0.

    so for our first root, x = b yields the minimum. for the second root, the same principle applies. we want to add 0 to c^2 in order for it to be a minimum. so we must have x = 0.

    so for y to be minimum, we need x = b or x = 0. how do we know which to choose? that depends on the size of a, b, and c
    Put a=1, b=2, c=3. Then plot the curve. Is the minimum at x=2 or x=0?
    (rhetorical question; no its not!)

    This curve is smooth and y can be minimised by solving dy/dx=0 in the usual
    manner (at least if it were not quite so nasty).

    RonL
    Last edited by CaptainBlack; March 17th 2007 at 12:49 AM.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Put a=1, b=2, c=3. Then plot the curve. Is the minimum at x=2 or x=0?
    (rhetorical question; no its not!)

    This curve is smooth and y can be minimised by solving dy/dx=0 in the usual
    manner (at least if it were not quite so nasty).

    RonL
    yeah, my bad, i guess i was trying to do the problem using as little calculus as possible
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  8. #8
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    Sorry about being rude before.. this assignment has come and past but i still want to know how to do these.. just for the sake of knowing something new.

    for the first problem.. how exactly would i solve dy/dx=0? (does it have something to do with "Implicit Differentiation"?)


    for the second problem.. i made a typo like jhevon said; it should be sinx+3 (NOT + 2). but yeah.. i need help with that as well.
    Last edited by tummbler; May 6th 2007 at 07:37 PM.
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