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Math Help - Minimisation Problem

  1. #1
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    Apr 2010
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    Minimisation Problem

    Hi, I have gotten halfway through this problem but I have gotten stuck. The problem is...

    The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials.

    I have done the following.

    Let the width = x, length = y and height = z; V (constant), V = xyz;

    Cost = 5xy + 2xz + 2yz
    Use V = xyz and sub z in for v/xy to get

    Cost = 5xy + 2V/y + 2V/x

    Minimizing this gives partial derivatives with respect to x and y

    Costx = 5y - 2V/(x^2) *1
    Costy = 5x - 2V/(y^2) *2

    let y = 2V/5(x^2) from *1, sub this into *2

    Giving...

    x^3 = 2V/5

    x = (2V/5)^1/3

    Now do we make y = x and then solve for the z dimension? This is where I am lost.

    Thanks in advance,

    Chris
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  2. #2
    MHF Contributor

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    Quote Originally Posted by purakanui View Post
    Hi, I have gotten halfway through this problem but I have gotten stuck. The problem is...

    The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials.

    I have done the following.

    Let the width = x, length = y and height = z; V (constant), V = xyz;

    Cost = 5xy + 2xz + 2yz
    Use V = xyz and sub z in for v/xy to get

    Cost = 5xy + 2V/y + 2V/x

    Minimizing this gives partial derivatives with respect to x and y

    Costx = 5y - 2V/(x^2) *1
    Costy = 5x - 2V/(y^2) *2

    let y = 2V/5(x^2) from *1, sub this into *2

    Giving...

    x^3 = 2V/5

    x = (2V/5)^1/3

    Now do we make y = x and then solve for the z dimension? This is where I am lost.
    Remember that you just said "let y = 2V/5(x^2) from *1". Solve for y from that.

    Thanks in advance,

    Chris
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