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Techniques of Differentiation: Applications of Derivatives

  1. Dec 29, 2011 #1
    1. The problem statement, all variables and given/known data

    We want to construct a box with a square base and we only have 10 m2 of material to use in construction of the box. Assuming that all the material is used in the construction process determine the maximum volume that the box can have.

    2. Relevant equations

    Chain rule. Second derivatives. Calculating Maximum and Minimum value.

    3. The attempt at a solution

    All the surface area = 10m2 = 2(x2)

    Base + Top + 4 vertical area = 10m2
    x2 + x2 + 4xy = 10

    y = (10 - 2x2) / 4x

    u = x2y

    r = x2((10 - 2x2) / 4x)

    du/dx = (5-3x2)/2

    -Okay, clearly I don't understand even a bit of my work. Someone please explain to me and show me the correct steps. Thanks-
     
  2. jcsd
  3. Dec 29, 2011 #2
    Look up the method of lagrange multipliers.
     
  4. Dec 29, 2011 #3
    What is that man. There suppose to be easy way to solve this, but I just can't see it. That Lagrange multipliers is not in my study's syllabus :|
     
    Last edited: Dec 29, 2011
  5. Dec 29, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Lagrange multipliers is too advanced for this problem.

    hadizainud, you almost have the problem solved. You used the surface area condition to write the volume as a function of x only and then differentiated with respect to x. I assume you did that because you know that "a function, f, has an extremum at:
    a point where f'(x)= 0.
    a point where f'(x) does not exist.
    and endpoint of the interval, if any.

    However, you have an error in your algebra:
    [tex]x^2\frac{10- 2x^2}{4x}= \frac{x^2(10- 2x^2)}{4x}= \frac{x(5- x^2)}{2}=\frac{5x- x^3}{2}[/tex]

    Find the derivative of that, set it equal to 0 and solve for x. What is the volume for that x?
     
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