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Math Help - Application of Differentiation..maximum and minimum values

  1. #1
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    Application of Differentiation..maximum and minimum values

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    Hello,..
    I can't find a solution for this question can anyone help plz..
    here is the question:

    A box is constructed out of a 3x4 m2 (metre square) rectangular piece of cardboard by cutting away at each edge a square of side length x and folding up each edge.

    (a) Find an expression of the box's volume in terms of x.
    (b) What are the possile values of x.
    (c) Determine the value of x which maximizes the volume of the box.
    Thanking you all in advance for helping...
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  2. #2
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    Quote Originally Posted by NiCeBoY View Post
    Note
    Hello,..
    I can't find a solution for this question can anyone help plz..
    here is the question:



    Thanking you all in advance for helping...
    Have you tried drawing a net for the box ....? Clearly the height of the box is x. What are the dimensions of the base .....?
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  3. #3
    Super Member 11rdc11's Avatar
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    You need to find the volume and then take the derivative of the volume and then equal the derivative to 0 which will give you the critical points.


    A=(3-2x)(4-2x)

    V=x(3-2x)(4-2x)

    \frac{dv}{dx} = 12x^2 - 28x + 12

    12x^2 -28x + 12 = 0
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  4. #4
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    on solving:
    12x^2 -28x + 12 = 0

    i get x=0.57 and x=1.78

    is this the possible values of x?

    and if x=1.78,is it when the voluma of the box is maximum plz.?

    thx
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  5. #5
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    x can't be greater than 1.5 ... can you reason why?
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  6. #6
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    yes because if one side of the card board is 3m and (2x1.5) should be remove so there will be nothing left..

    so how i should proceeed plz.?
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  7. #7
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    you have another solution ... how can you check that it yields a maximum?
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  8. #8
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    when equating dv/dx with o if the value of x is negative its maximum and its positive its a minimum.. i think..
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  9. #9
    Super Member 11rdc11's Avatar
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    Try using the second derivative test to see if it a max
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    when i equate the second derivative of :
     12x^2 -28x + 12

    which is

     24x - 28

    to 0 the value of x is 1.17 which is positive. so i think it should be a minimum..
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  11. #11
    Super Member 11rdc11's Avatar
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    Try again 1.17 is a point of inflection. You on the right track but instead of equaling the second derivative to 0 just plug in your critical point from the first derivative into the second derivative. If it gives you a positive answer it a minimum and if produces a negative answer it is a maximum.
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  12. #12
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    that is i use the value of x=0.57 and x=1.78 and see if its positive or negative? am i right?
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  13. #13
    Super Member 11rdc11's Avatar
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    You can't use 1.78 for the reason you stated earlier. Use .57


    -14.32 = 24(.57) - 28

    The 2nd derivative gives a negative answer so it is a maximum.

    Hope this helps.
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  14. #14
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    Then how do i answer (b) What are the possile values of x.

    i think the answer is : 0< X < 1.5

    Am i rite plz?
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  15. #15
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NiCeBoY View Post
    Then how do i answer (b) What are the possile values of x.

    i think the answer is : 0< X < 1.5

    Am i rite plz?
    yup
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