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Math Help - Min/Max Surface Area Cubes (chk anwser)

  1. #1
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    Min/Max Surface Area Cubes (chk anwser)

    A mass of clay of volume 432 in.^3 is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

    (A) The minimum total surface area is obtained by making one cube as small as possible, .01^3, and the second cube 7.5595^3. The combined total surface area of these two cubes is (0.06+342.8) = 342.86.

    The maximum total surface area can be reached with two equal cubes. 6^3, for each cube. The combined total surface area of these two cubes is (216^2) = 432.
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  2. #2
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    Quote Originally Posted by Yogi_Bear_79
    A mass of clay of volume 432 in.^3 is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

    (A) The minimum total surface area is obtained by making one cube as small as possible, .01^3, and the second cube 7.5595^3. The combined total surface area of these two cubes is (0.06+342.8) = 342.86.

    The maximum total surface area can be reached with two equal cubes. 6^3, for each cube. The combined total surface area of these two cubes is (216^2) = 432.
    A much different approach:

    volume of a cube is: s^3

    surface area of a cube is: 6s^2

    so then we get V_1+V_2=432

    so then we say that: V_2=432-V_1

    then rewrite: s_2^3=432-s_1^3

    therefore: s_2=\sqrt[3]{432-s_1^3}

    we're trying to find surface area (SA) so we say: SA=SA_1+SA_2

    therefore: SA=6s_1^2+6\sqrt[3]{432-s_1^3}

    you can then evaluate the answers from that.
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    Hello, Yogi_Bear_79!

    Since this is posted under Calculus, I assume we're supposed to do Calculus.


    A mass of clay of volume 432 in³ is formed into two cubes.
    What is the minimum possible total surface area of the two cubes?
    What is the maximum?

    Let x and y be the sides of the two cubes.

    Their volumes satisfy: . x^3 + y^3\:=\:432\quad\Rightarrow\quad y^3\:=\:432- x^3\quad\Rightarrow\quad y^2\:=\:\left(432-x^3\right)^{\frac{2}{3}} [1]

    Their total surface area is: . A\;=\;6x^2 + 6y^2 [2]

    Substitute [1] into [2]: . A \;= \;x^3 + 6\left(432 - x^3\right)^{\frac{2}{3}}

    Differentiate: . A' \;= \;12x + 6\cdot\frac{2}{3}\left(432-x^3\right)^{-\frac{1}{3}}(-3x^2)

    And we have: . 12a - \frac{12x^2}{(432-x^3)^{\frac{1}{3}}} \;=\;0

    Multiply by (432 - x^3)^{\frac{1}{3}}:\;\;12x(432 - x^3)^{\frac{1}{3}} - 12x^2\;=\;0

    Factor: . 12x\left[(432 - x^3)^{\frac{1}{3}} - x\right] \;= \;0


    We have two equations to solve:

    . . 12x \,= \,0\quad\Rightarrow\quad \boxed{x \,= \,0}

    . . (432-x^3)^{\frac{1}{3}} - x\:=\:0\quad\Rightarrow\quad(432-x^3)^{\frac{1}{3}}\:= \:x\quad\Rightarrow\quad 432 - x^3\:=\:x^3

    . . . . . . . . \Rightarrow\quad 2x^3\:=\:432\quad\Rightarrow\quad x^3\:= \:216\quad\Rightarrow\quad \boxed{x\,=\,6}


    x = 0 means that the clay is formed into one cube.
    . . Its surface area is: . A\:=\:6(432)^{\frac{2}{3}} \:\approx\:342.88 in² . . . minimum

    x = 6 means the clay is formed into two equal cubes.
    . . Their surface area is: . A\:=\:6^3 + 6(216)^{\frac{2}{3}}\:=\:432 in² . . . maximum

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    Here's the graph of the equation:
    (it looks as though there are 2 minimum values)
    Attached Thumbnails Attached Thumbnails Min/Max Surface Area Cubes (chk anwser)-derivatives.jpg  
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  5. #5
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    Thank you both for your help. I have a question. The original question states that the clay is formed into two cubes. So wouldn't that make Sorban's minimum anwser incorrect for the question?
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    Hello again, Yogi_Bear_79!

    The original question states that the clay is formed into two cubes.
    So wouldn't that make Sorban's minimum answer incorrect for the question?

    I thought about that . . . and had to make a decision.

    If one cube is infinitesimally small, the total surface area approaches 6(432)^{\frac{2}{3}}
    . . Hence, there is no minimum surface area.

    So I assumed that one cube of side 0 is allowed.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    There are two minimums because there are two solutions:
    . . (x,y) \,=\,(0,6) and (6,0).

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    It makes more sense to say that there is not minimum because the interval is open. And thus you do not include the endpoints.
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    Quote Originally Posted by Yogi_Bear_79
    A mass of clay of volume 432 in.^3 is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

    (A) The minimum total surface area is obtained by making one cube as small as possible, .01^3, and the second cube 7.5595^3. The combined total surface area of these two cubes is (0.06+342.8) = 342.86.

    The maximum total surface area can be reached with two equal cubes. 6^3, for each cube. The combined total surface area of these two cubes is (216^2) = 432.
    There is a minimum allowed length of .01 of the side of a cube. So two cubes are required, the allowed lengths form a closed interval and a minimum exists. But that requires some change to Soroban's proof to show that the minimum surface area occurs at an endpoint. Quick's graph shows this but some proof is probably necessary for the exercise.
    Last edited by JakeD; August 8th 2006 at 08:20 PM.
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