Calculus (Optimization)

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A rectangular piece of cardboard, 8 inches by 14 inches, is used to make an open top box by cutting out a small square from each corner and bending up the sides. What size square should be cut from each corner for the box to have the maximum volume?
So far I have: V = (14 - 2x)(8 - 2x)(h)

  • Calculus (Optimization, Still Need Help) -

    I just wanted to correct something for my equation, it should be: V = (14 - 2x)(8 - 3x)(x), which simplifies to V = 112x - 44x^2 - 4x^3.
    Take the derivative:
    V' = 112 - 88x - 12x^2
    Now all I need are the roots, any help? I think I found one around 1.10594, possibly?

  • Calculus (Optimization) -

    set V'=0, and you have a quadratic. Why not use the quadratic equation..

    12x^2+88x-112=0
    3x^2+22x-28=0

    x= (-22+-sqrt (22^2+12*28))/6

    doing it in my head, I get about..

    (-22+-28)/6= 4/6, 7.5 in my head.

    check my work and estimates.

  • Calculus (Optimization) -

    I think that you might have gotten the equation wrong, I think that it should be: 3x^2 - 22x + 28. When I put this equation into the quadratic equation, I got 5.694254177 and 1.639079157. So the squares that need to be cut out should have an area of approximately 2.69 square inches?

  • Calculus (Optimization) -

    I don't see how you got the signs as you did. Please recheck

  • Calculus (Optimization) -

    hang on, I reread the problem statement. In the first response I gave, I took your equation. I don't think it is right.
    give me a minute.

  • Calculus (Optimization) -

    Ok, your equation is right. Recheck your final signs as I stated.

  • Calculus (Optimization) -

    I rechecked and found that 3x^2-22x+28 has the correct signs. Knowing this equation and the values I found from the quadratic equation, would you say that the 1.639079157 term is correct? (The 2.69 square inches came from squaring the 1.639079157).

  • Calculus (Optimization) -

    Let me do some thinking...
    if 0=V' = 112 - 88x - 12x^2
    multipy both sides by -1, and
    12x^2+88x-112=0

    I don't see those as your signs....

  • Calculus (Optimization) -

    Now I'm lost, I don't get why you changed the signs.

  • Calculus (Optimization) -

    You are lost. This is algebra.

    if 0=112 - 88x - 12x^2
    do whatever you know to put it in standard form, ax^2+bx+c=0
    when you do that a and b will have the SAME signs. Surely you can do that.
    if a=-12, then b=-88, and c=112
    if a=12, then b=88, and c=-112

  • Calculus (Optimization) -

    Okay, so does this change my original answer of approximately 1.64 to 4.42??? The 4.42 came from putting your new values in the quadratic equation.

  • Calculus (Optimization) -

    Nevermind, that 4.42 was a mistake and my very original answer of 1.105940354 was absolutely correct!!! This is the right answer, I know it!

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