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Thread: Cost Optimization

  1. #1
    MIN
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    Cost Optimization

    Hey guys could you please help me with this question??

    ____________
    In the year 2010, the drought in Melbourne resulted in water restrictions across the metropolitan area. In order to maintain a fresh water supply for my fish tank, I decided to construct a 72m^3 open-topped, rectangular rainwater
    tank in my backyard. The available tanks all had square bases, with side length x and height y . The smalles


    size had a 1m^2 base and due to space constraints, the height of the tank could be no more than 9/8m.
    The cost of the tank plus installation is given by c(x) =5(x^2+4xy)+10xy


    what shape did I make the tank ( that is, what were the values of x and y) in order to minimize the cost?
    _____________

    The answer says I'm supposed to get 8 for x value but I'm not getting that..
    since equation for volume is x^2*y I set that equal to 75 and arranged it to get y= 72/x^2 and I substituted that to the cost equation 5(x^2+4xy)+10xy,
    differentiated it and set it equal to 0. Am I doing something wrong??
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  2. #2
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    Re: Cost Optimization

    I don't agree with x = 8m ... I get x = 6m and y = 2m


    oh, and what is this significance of this cryptic piece of info?

    The smalles


    size had a 1m^2 base and due to space constraints, the height of the tank could be no more than 9/8m
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  3. #3
    MIN
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    Re: Cost Optimization

    yeah I was wandering the same thing... it's saying that y can't be greater than 9/8 m so 2 doesn't work i guess..
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    Re: Cost Optimization

    If the height can be no greater than $\dfrac{9}{8} \, m$, then $x^2y = x^2 \cdot \dfrac{9}{8} = 72 \implies x = 8 \, m$
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