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Math Help - Help getting started with a couple optimizations

  1. #1
    Junior Member NAPA55's Avatar
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    Help getting started with a couple optimizations

    Sometimes I can do these... sometimes I look at them and don't know where to begin.

    If someone could give me a couple tips to get these few started, I can do the rest:

    1. A box with a square base and no top must have a volume of 10 000 cm3. If the smallest dimension in any direction is 5 cm, then determine the dimensions of the box that minimize the amount of material used.

    2. A fence is 1.5 m high and is 1 m from a wall. A ladder must start from the ground, touch the top of the fence, and rest somewhere on the wall. Find the minimum length of such a ladder.

    3. The motion of a particle is given by s(t) = 5cos(2t + pi/4). What are the maximum values of the displacement, the velocity, and the acceleration?
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by NAPA55 View Post
    Sometimes I can do these... sometimes I look at them and don't know where to begin.

    If someone could give me a couple tips to get these few started, I can do the rest:

    1. A box with a square base and no top must have a volume of 10 000 cm3. If the smallest dimension in any direction is 5 cm, then determine the dimensions of the box that minimize the amount of material used.

    2. A fence is 1.5 m high and is 1 m from a wall. A ladder must start from the ground, touch the top of the fence, and rest somewhere on the wall. Find the minimum length of such a ladder.

    3. The motion of a particle is given by s(t) = 5cos(2t + pi/4). What are the maximum values of the displacement, the velocity, and the acceleration?
    For the first one the surface area is given by the formula

    \underbrace{x^2}_{bottom}+\underbrace{4xy}_{sides}  =10,000 \mbox{ and } V=x^2y

    Solve the first for y and sub into the second and minimize the function

    V=2500x-\frac{x^3}{4}
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Heres three

    the first two are fairly simple f(t)=5cos\bigg(2t+\frac{\pi}{4}\bigg)...then v(t)=f'(t)=-5sin\bigg(2t+\frac{\pi}{4}\bigg)\cdot{2}...set that equal to zero you get...wait there is no boundary or artificial cut off...well I am just going to assume it is (0,2\pi) the values are x=\frac{3\pi}{8},x=\frac{7\pi}{8}.x=\frac{11\pi}{8  },x=5.89 next put those values in the second derivative...the negative values are maxs and the minimums are negative...repeat for the other two with the derviative...solving and putting that in its derivative
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    similar triangles

    We need to use similar triangles to find d in the diagram below.(sorry without the buttons I don't know how to add it in the middle of a post)

    \frac{d}{1.5}=\frac{x+1}{x} \iff d=\frac{3(x+1)}{2x}

    Since we want the smallest length of the latter we need to know the length of the hypotenuse. By the pythagorean theorem we get

    h=\sqrt{\left( \frac{3(x+1)}{2x}\right)^2+(x+1)^2}

    You just need to minimize from here.

    Good luck.
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  5. #5
    Junior Member NAPA55's Avatar
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    Thank you both!
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