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Math Help - Expected cost of making dice

  1. #1
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    Expected cost of making dice

    Sorry for the wordy problem.. just not sure if i have done this correctly;

    A straight wooden plank of length 2 m has been purchased for $34. This plank is about to be sawed into two at a point chosen randomly along it. Each of the two resulting short planks will then be sawed lengthwise into twelve identical sticks. Next, two perfectly cubic frames will be constructed from the twenty four sticks. Then both frames will be made into huge dice by covering them, each on all six sides, with a type of red plastic sheet which costs $42 per square metre. Next, each die will be filled with a bouncy rubber material which costs $66 per cubic metre. Finally, the
    usual number of dots will be put onto the dice. For each die whose total surface area is at least 5400 square centimetres, the dots will be made of gold at a cost of $15 each; for each die smaller than this, the dots will be painted on with black paint at negligible cost.
    For how many dollars should the two dice be sold (in advance) so that the expected profit is $700?

    So letting the length from the end of the plank to the cut being x to form the first dice, and from the cut to the other end 2-x for the second dice, so x is uniformally distributed along the plank, and  E(X)=\frac{2-1}{2}=1
    So surface areas and volumes of the boxes are
    SA_{1}=6\cdot (X)^2
    SA_{2}=6\cdot (2-X)^2
    V_{1}=x^3
    V_{2}=(2-x)^3

    And for the dots, we know that if SA=6\cdot (X)^2\geq 5.4m^2 then we need to buy the expensive dots, so if X\geq .3 the first box will have a surface area that requires us to buy the dots, and if X\leq 1.7 the second box will also require dots.. So the expected cost of the dots is
    E(\mbox{Cost of dots})=E(\mbox{1 dice needs dots})+E(\mbox{both dice needs dots})=\frac{0.3+0.3}{2} \cdot 15\cdot 21+\frac{1.4}{2}\cdot 15\cdot 21\cdot 2=\$535.50

    And the expected cost of the dice is
    E(C)=34+42\cdot \left[E(X)\right]^{2}+42\cdot \left[E(2-X) \right]^{2}+66\cdot  \left[E(X)\right]^{3} + 66\cdot  \left[E(2-X)\right]^{3}+E(\mbox{cost of dots}) =34+252\cdot 1^2 + 252\cdot (2-1)^2+ 66\cdot 1^3 + 66\cdot (2-1)^2 + 535.50=\$1205.5

    So since we want E(\mbox{Profit})=\$700; The Price should be set to P=E(\mbox{Profit})+E(\mbox{Cost})=700+1205.5=\$190  5.5

    Thanks for any help
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Robb View Post
    Sorry for the wordy problem.. just not sure if i have done this correctly;

    A straight wooden plank of length 2 m has been purchased for $34. This plank is about to be sawed into two at a point chosen randomly along it. Each of the two resulting short planks will then be sawed lengthwise into twelve identical sticks. Next, two perfectly cubic frames will be constructed from the twenty four sticks. Then both frames will be made into huge dice by covering them, each on all six sides, with a type of red plastic sheet which costs $42 per square metre. Next, each die will be filled with a bouncy rubber material which costs $66 per cubic metre. Finally, the
    usual number of dots will be put onto the dice. For each die whose total surface area is at least 5400 square centimetres, the dots will be made of gold at a cost of $15 each; for each die smaller than this, the dots will be painted on with black paint at negligible cost.
    For how many dollars should the two dice be sold (in advance) so that the expected profit is $700?

    So letting the length from the end of the plank to the cut being x to form the first dice, and from the cut to the other end 2-x for the second dice, so x is uniformally distributed along the plank, and  E(X)=\frac{2-1}{2}=1
    So surface areas and volumes of the boxes are
    SA_{1}=6\cdot (X)^2
    SA_{2}=6\cdot (2-X)^2
    V_{1}=x^3
    V_{2}=(2-x)^3

    And for the dots, we know that if SA=6\cdot (X)^2\geq 5.4m^2 then we need to buy the expensive dots, so if X\geq .3 the first box will have a surface area that requires us to buy the dots, and if X\leq 1.7 the second box will also require dots.. So the expected cost of the dots is
    E(\mbox{Cost of dots})=E(\mbox{1 dice needs dots})+E(\mbox{both dice needs dots})=\frac{0.3+0.3}{2} \cdot 15\cdot 21+\frac{1.4}{2}\cdot 15\cdot 21\cdot 2=\$535.50

    And the expected cost of the dice is
    E(C)=34+42\cdot \left[E(X)\right]^{2}+42\cdot \left[E(2-X) \right]^{2}+66\cdot  \left[E(X)\right]^{3} + 66\cdot  \left[E(2-X)\right]^{3}+E(\mbox{cost of dots}) =34+252\cdot 1^2 + 252\cdot (2-1)^2+ 66\cdot 1^3 + 66\cdot (2-1)^2 + 535.50=\$1205.5

    So since we want E(\mbox{Profit})=\$700; The Price should be set to P=E(\mbox{Profit})+E(\mbox{Cost})=700+1205.5=\$190  5.5

    Thanks for any help
    That looks OK (except for the typo SA=6\cdot (X)^2\geq 5.4m^2 instead of SA=6\cdot (X)^2\geq 0.54m^2)

    CB
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  3. #3
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    Why is X^2 used in SA_1 rather than  \frac{1}{12^2} X^2 (the length of one side of the cube? Or am I not understanding something here?

    Cheers,

    Adam
    Last edited by theBFG; October 20th 2009 at 04:45 AM. Reason: typo
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by theBFG View Post
    Why is X^2 used in SA_1 rather than  \frac{1}{12^2} X^2 (the length of one side of the cube? Or am I not understanding something here?

    Cheers,

    Adam
    The original timber is cut at X, then sawn longitudinally into 12 pieces all of length X, similarly for the other piece.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    The original timber is cut at X, then sawn longitudinally into 12 pieces all of length X, similarly for the other piece.

    CB
    Oops! Had misinterpretted "lengthwise"... Thanks for your help.
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