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Math Help - Need help with 2 questions

  1. #1
    Super Member 11rdc11's Avatar
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    Need help with 2 questions

    Hi guys for some reason I'm having problems with these 2 questions, Any help would be great.

    1. For the equation 60x^3-3x^5 = y where does the largest slope occur?

    Ok here is how I thought I would find the answer for this problem. If I take the second derivative and equal it to 0 and find my critical points, then one of my points will be the the largest slope. Is this correct?

    f(x) = 60x^3-3x^5

    f'(x) = 180x^2 -15x^4

    f''(x) = 360x -60x^3

    so

    0 = 60x(6-x^2)

    so

    x = 0, \sqrt{6}, -\sqrt{6}

    so I think my largest slope occurs at

    \sqrt{6}, -\sqrt{6}

    Is this correct?

    2.A rectangular storage container with a top is to have a volume of 6 M^3. The length of the base is 1.5 times the width. Material for the base costs $3 per square meter, material for the sides cost $5 per square meter and the material for the top costs $8 per square meter. Find the dimensions of the container with minimal total cost.


    I'm not sure how to do the second one.

    I know

    w=x
    l=1.5

    but what do I use for height?

    Also I know I have to use surface area but can't come up with an equation without the height.
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  2. #2
    Super Member 11rdc11's Avatar
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    Ok so I tried this approach but not sure if it right

    w=x
    l = 1.5x
    h=y

    so for volume

    6 = (x)(1.5x)(y)

    Then I solve for y

    y = 4x^2

    SA = 2(1.5x)(x) +2(1.5x)(y) + 2(x)(y)

    Then sub y in

    SA= 2(1.5x^2) +2(1.5x)(4x^2) +2(x)(4x^2)

    3x^2 +20x^3

    Then I'm lost after that lol
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Ok so I tried this approach but not sure if it right

    w=x
    l = 1.5x
    h=y

    so for volume

    6 = (x)(1.5x)(y)

    Then I solve for y

    y = 4x^2

    SA = 2(1.5x)(x) +2(1.5x)(y) + 2(x)(y)

    Then sub y in

    SA= 2(1.5x^2) +2(1.5x)(4x^2) +2(x)(4x^2)

    3x^2 +20x^3

    Then I'm lost after that lol
    should it not be y = \frac 4{x^2}?

    and remember, you want the cost equation, not the surface area equation to be your objective

    after you get that, just find the minimum points of that function
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Jhevon View Post
    should it not be y = \frac 4{x^2}?

    and remember, you want the cost equation, not the surface area equation to be your objective

    after you get that, just find the minimum points of that function
    lol yep it should be

    y =\frac{4}{x^2}

    So the derivative of the SA is the Cost function? If not how do I set up the cost function?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    lol yep it should be

    y =\frac{4}{x^2}

    So the derivative of the SA is the Cost function? If not how do I set up the cost function?
    the cost for each "side" (can mean top or bottom) of the box is (cost of the side)*(area of the side)

    think you can find the equation now?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Hi guys for some reason I'm having problems with these 2 questions, Any help would be great.

    1. For the equation 60x^3-3x^5 = y where does the largest slope occur?

    Ok here is how I thought I would find the answer for this problem. If I take the second derivative and equal it to 0 and find my critical points, then one of my points will be the the largest slope. Is this correct?

    f(x) = 60x^3-3x^5

    f'(x) = 180x^2 -15x^4

    f''(x) = 360x -60x^3

    so

    0 = 60x(6-x^2)

    so

    x = 0, \sqrt{6}, -\sqrt{6}

    so I think my largest slope occurs at

    \sqrt{6}, -\sqrt{6}

    Is this correct?
    by the way, this was correct. these are the x-values where you have the maximum slope.
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Jhevon View Post
    by the way, this was correct. these are the x-values where you have the maximum slope.
    Thanks so the x values of \sqrt{6}, - \sqrt{6}, but not the x value of 0 right?

    The cost function would be

    C = 4.5x^2 +12x^2 +\frac{60}{x} + \frac{40}{x}

    correct?

    Thanks for the help
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Thanks so the x values of \sqrt{6}, - \sqrt{6}, but not the x value of 0 right?
    that's right. x = 0 is a local minimum for the derivative function.

    The cost function would be

    C = 4.5x^2 +12x^2 +\frac{60}{x} + \frac{40}{x}

    correct?

    Thanks for the help
    yes, that's correct. you can write it as C = 16.5x^2 + \frac {100}x though

    now find the minimum points of that function. you will be able to use that to find the dimensions
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  9. #9
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Jhevon View Post
    that's right. x = 0 is a local minimum for the derivative function.

    yes, that's correct. you can write it as C = 16.5x^2 + \frac {100}x though

    now find the minimum points of that function. you will be able to use that to find the dimensions
    Thanks that was my next step.
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