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Math Help - Solving word problems using derivatives

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    Solving word problems using derivatives

    I have four word problems that are due tomorrow, and my main problem is that I can't figure out how to get the formulas. We have to somehow end up with a quadratic that we can get a first and second derivative from in order to find a max or min. If someone could at least help out with the formulas, I should be able to solve the rest of the problem. I especially don't understand 2 and 4, so if anyone could try to explain them, it would be much appreciated.

    1. A chartering company will provide a plane for a fare of $300 per person for 100 or fewer passengers (but more than 60 passengers.) For each passenger over 100, the fare is decreased $2 per person for everyone. What number of passengers would provide the greatest revenue?

    2. At noon, ship A, steaming east at 10 knots, is 130 nautical miles due north of ship B, which is steaming north at 15 knots. When will the ships be closest together?

    3. A box of volume 32 cubic feet having a square base and no top is to be constructed from materials costing $1.00 per square foot. Find the dimensions of such a box for which the cost of material is least.

    4. Find the point on the parabola y=x^2 that is closest to the point P(3,0).
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    Re: Solving word problems using derivatives

    The most critical part of such problems is determining the function which represents the given model. It is extremely important to be able to find these on your own.

    1.) Revenue is price per passenger times the number of passengers. For this one you need a piecewise defined function. What is your independent variable, and what are the intervals over which we need to make definitions?
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    Re: Solving word problems using derivatives

    I typed the entire question up there; that's all my teacher gives us. The only problem with that, is that I don't know how to find a max on a piecewise function. The formula for 60-100 passengers would be 300x, and for anything greater than 100 would be (300-2x)(100+x), correct?
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    Re: Solving word problems using derivatives

    You have the correct revenue function for 60<x\le100 but not for 100<x.

    The fair per passenger begins at $300 where x = 100, then decreases by $2 for each additional passenger. This is a linear relationship, and we know a point (100,300) and the slope ( m=-2), so use the point-slope formula to determine the fair per passenger. What do you find as the fair per passenger for 100<x?

    Note: the number of passengers over 100 is 100-x.
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    Re: Solving word problems using derivatives

    y=-2x+500?
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    Re: Solving word problems using derivatives

    Yes, good work! So what is the revenue function then over the relevant interval?
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    Re: Solving word problems using derivatives

    y=-2x+500, 60<x\le100?
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    MHF Contributor MarkFL's Avatar
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    Re: Solving word problems using derivatives

    Sorry for the delay, I was battling with LaTeX do display a piecewise function to no avail...

    No, that is the fair per passenger, you need to multiply y by x (the number of passengers) to get the revenue. For example, when 60<x\le100 we have R(x)=300x since the fair per passenger is 300 and there are x passengers. Likewise we need R(x)=(500-2x)x=500x - 2x^2 when 100<x since the fair per passenger is 500-2x and there are x passengers.

    Does this make sense?
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    Re: Solving word problems using derivatives

    Somewhat... I'll try to finish it and see if the answer works out. Thank you.
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    Re: Solving word problems using derivatives

    Wait. I have two functions now, but I'm not allowed to use a calculator. We're supposed to find the first derivative (I can do that), but do I need to combine those two equations?
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    Re: Solving word problems using derivatives

    In the last question, you are trying to MINIMIZE distance. Think, what formula do you know that can determine the distance between two points on the coordinate plane? Then, find your substitution for y, differentiate, and solve for x. Then, find the point you are looking for by plugging in the x you found and then solve for y to find the point you are looking for.
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    Re: Solving word problems using derivatives

    You just gave me a headache... how exactly do I differentiate, and how do I know what to substitute in for y?
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    Re: Solving word problems using derivatives

    edit: referring to the first problem:

    You need to analyze them separately over their respective domains. For example, we know in the leftmost domain, the maximum revenue occurs at the right end-point, since the revenue is linear and has positive slope. So, then we are left to look at the other piece, which has a quadratic revenue function, and can actually be tackled with pre-calculus techniques. We know we have a parabola opening downward, and so the maximum will occur at the axis of symmetry.

    I would use calculus though, since this is the forum you posted in, so I assume you are supposed to apply the techniques of differential calculus to optimize the revue function. It is nice though to have an alternate method with which to check your work.

    Let us know what you find.
    Last edited by MarkFL; January 6th 2013 at 12:06 PM.
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    Re: Solving word problems using derivatives

    Sorry. Didn't want to give much away without me doing the problem for you.

    Use the distance formula. Your substitution needed is y=x^2. Once you have everything in terms of x, differentiate.
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    Re: Solving word problems using derivatives

    Nevermind. My teacher never told us that finding derivatives was called differentiation. I feel like quite the idiot now.
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