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Math Help - Calculus

  1. #1
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    Calculus

    1. How close does the semicircle x^2 + y^2 = 16 come close to the point (3^1/2, 1)? (Hint: if you minimize the square of the distance, you can avoid dealing with square roots.)

    2. The U.S. Postal Service wil accept a box for domestic shipment only if the some of its length and girth (girth being total distance around the package once) does not exceed 92 inches. What dimensions will give a box with a square end the largest possible volume?

    3. The Strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a 12-inch diameter log.

    I am in the first semester of calculus in college, so this work would have to be done with that, can anyone help me? I don't understand this at all.
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    Quote Originally Posted by Prophet View Post
    1. How close does the semicircle x^2 + y^2 = 16 come close to the point (3^1/2, 1)? (Hint: if you minimize the square of the distance, you can avoid dealing with square roots.)
    you want to minimize the distance between the points \left( x, \pm \sqrt{16 - x^2} \right) and ( \sqrt{3},1)

    use the distance formula, then square both sides. minimize the resulting function
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    Quote Originally Posted by Prophet View Post
    2. The U.S. Postal Service wil accept a box for domestic shipment only if the some of its length and girth (girth being total distance around the package once) does not exceed 92 inches. What dimensions will give a box with a square end the largest possible volume?
    see here
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  4. #4
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    #3.

    We are told the strength is proportional to the product of the width and the square of the height(depth).

    So, we can write S=kwh^{2}

    S is the strength and k is a constant that depends on the wood type, etc.

    We find a function that relates w and h. Draw a line from the center of the log to the corner of the beam.

    We get a right triangle. (\frac{w}{2})^{2}+(\frac{h}{2})^{2}=36

    w^{2}+h^{2}=144

    h^{2}=144-w^{2}

    Now, sub into the strength equation:

    S=kw(144-w^{2})=k(144w-w^{3})

    Now, differentiate, set to 0 and solve for w. h will follow.
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    Quote Originally Posted by Prophet View Post
    3. The Strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a 12-inch diameter log.
    Let S be the strength of the beam
    Let w be the width of the beam
    Let d be the depth of the beam
    Let k be the proportionality constant

    Since the strength is proportional to the width times the square of the depth, we have that:

    S = kwd^2

    thus, to maximize the strength, we need to maximize the dimensions of the rectangle.

    now, imagine the end of the log as a circle centered at the origin. the equation of this circle is therefore given by w^2 + d^2 = 36 ....(i use w on the x-axis and d on the y-axis)

    we want to maximize the rectangle we can get in the upper half of the circle. draw a diagram and you will realize that the length of the rectangle is given by 2w and the height (or depth) is given by d

    thus we have the Area = height*width being A = 2wd

    let's maximize this. our constraint is that d = \sqrt{36 - w^2}

    thus we have A = 2w \left( \sqrt{36 - w^2} \right) now find the values for w and d that maximizes this. (find A' and set it to zero and solve for w, then plug the value you got for w into the original equation to solve for d)

    (alternatively, which may be more appropriate for this question, you can maximize the strength directly by plugging in d = \sqrt{36 - w^2} into the equation for S. yes, i think that would be better)
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    Quote Originally Posted by galactus View Post
    #3.

    We are told the strength is proportional to the product of the width and the square of the height(depth).

    So, we can write S=kwh^{2}

    S is the strength and k is a constant that depends on the wood type, etc.

    We find a function that relates w and h. Draw a line from the center of the log to the corner of the beam.

    We get a right triangle. (\frac{w}{2})^{2}+(\frac{h}{2})^{2}=36

    w^{2}+h^{2}=144

    h^{2}=144-w^{2}

    Now, sub into the strength equation:

    S=kw(144-w^{2})=k(144w-w^{3})

    Now, differentiate, set to 0 and solve for w. h will follow.
    i like your right-triangle solution. it seems we would end up with different solutions though. i figure mine is incorrect, since i went through a lot more stuff to get to the meat of the matter. i likely made an error somewhere along the long journey
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    Quote Originally Posted by Jhevon View Post
    you want to minimize the distance between the points \left( x, \pm \sqrt{16 - x^2} \right) and ( \sqrt{3},1)

    use the distance formula, then square both sides. minimize the resulting function
    Haha :P How do you minimize the function?
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    Quote Originally Posted by Prophet View Post
    Haha :P How do you minimize the function?
    find it's derivative and set it equal to zero and solve for the variable

    you remember the distance formula right?
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    Quote Originally Posted by Jhevon View Post
    find it's derivative and set it equal to zero and solve for the variable

    you remember the distance formula right?
    Yeah thanks.
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    So I plugged it in the formula... and then I squared both sides... but if I take the derivative, how do I work with the left side... the F^2 ???
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    Quote Originally Posted by Prophet View Post
    So I plugged it in the formula... and then I squared both sides... but if I take the derivative, how do I work with the left side... the F^2 ???
    just call it something else. so if we call D the distance formula, then the formula you want is that of D^2. but let this be a variable unto itself, that is, call D^2 a new variable, say F. so the right of your equation is now F only.
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    Quote Originally Posted by Jhevon View Post
    just call it something else. so if we call D the distance formula, then the formula you want is that of D^2. but let this be a variable unto itself, that is, call D^2 a new variable, say F. so the right of your equation is now F only.
    So with that information taken in... I have F^2 = D and so
    D' = 2~3^(1/2) - 2x - 2x(16-x^(2))^(-1/2) is this right??? then I set it equal to zero?
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    Quote Originally Posted by Prophet View Post
    So with that information taken in... I have F^2 = D and so
    D' = 2~3^(1/2) - 2x - 2x(16-x^(2))^(-1/2) is this right??? then I set it equal to zero?
    Seems a bit complicated :P.
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  14. #14
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    Quote Originally Posted by Prophet View Post
    So with that information taken in... I have F^2 = D and so
    D' = 2~3^(1/2) - 2x - 2x(16-x^(2))^(-1/2) is this right??? then I set it equal to zero?
    i'm not sure what you are doing here. to be clear, we want to minimize the function

    D = (\sqrt{3} - x)^2 + \left(1 - \sqrt{16 - x^2} \right)^2
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    Quote Originally Posted by Jhevon View Post
    i'm not sure what you are doing here. to be clear, we want to minimize the function

    D = (\sqrt{3} - x)^2 + \left(1 - \sqrt{16 - x^2} \right)^2

    So, then you take the derivative of this function and then set it equal to 0, I took the derivative and got that function... but I was just wondering if my differentiation was correct.
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