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Thread: Word/equation question

  1. #1
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    Word/equation question

    I have done questions like this before, But i am not sure how to start this one. Once I have the beginning equation I should be able to do it. As for the second one we havent done much on this in class so I'm not sure where to start.
    Any help would be great. Thanks
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  2. #2
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    Quote Originally Posted by stacyg View Post
    I have done questions like this before, But i am not sure how to start this one. Once I have the beginning equation I should be able to do it. As for the second one we havent done much on this in class so I'm not sure where to start.
    Any help would be great. Thanks
    1. If the total area is a = 3000 cm and the width of the poster is x then you know:

    $\displaystyle l = \dfrac{3000}x\ ,\ w = x$

    2. Cut of the margin:
    from the length cut of 12 cm in total and from the width cut of 4 cm in total. The area of the printing part is then:

    $\displaystyle a_p = \left(\dfrac{3000}x - 12 \right) (x - 4) = -12x - \dfrac{12000}x + 3048$

    3. Now calculate the maximum of $\displaystyle a_p$ using derivation.

    --------------------------------

    You are supposed to know that $\displaystyle speed = \dfrac{distance}{time}~\implies~time = \dfrac{distance}{speed}$

    Therefore: Traveling time is $\displaystyle t = \dfrac{60}v$

    Total costs: $\displaystyle C(v)= \underbrace{\left(\dfrac{60}v \right)}_{time}\cdot \overbrace{\left(\dfrac12 \cdot v^3 + 20 \underbrace{+ 3000}_{fix \ costs} \right)}^{costs\ per\ hour}$

    Now calculate the maximum of $\displaystyle C(v)$ using derivation.
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  3. #3
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    Thanks. I'm not sure if I've done the differentiation right but for the 1st question would x=12 therefore the area be 1904cm squared.

    And for the second question would c(v)=150.

    Thanks
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  4. #4
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    Quote Originally Posted by stacyg View Post
    Thanks. I'm not sure if I've done the differentiation right but for the 1st question would x=12 therefore the area be 1904cm squared.

    And for the second question would c(v)=150.

    Thanks
    1. If the width of the "poster" is 12 cm then the length must be 250 cm, that means 2 and a half meter. That's a quite unusual format for a poster.
    2. With C(v) = 150 you can pay 3 minutes of the fixed costs!

    3. As you suspected there must be gone something wrong. Therefore I'll give you the equations which you must solve for x or v respectively:

    $\displaystyle \dfrac{12000}{x^2} - 12 =0$


    $\displaystyle 60v-\dfrac{181200}{v^2} = 0$

    Good luck!
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  5. #5
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    Quote Originally Posted by earboth View Post
    1. If the width of the "poster" is 12 cm then the length must be 250 cm, that means 2 and a half meter. That's a quite unusual format for a poster.
    2. With C(v) = 150 you can pay 3 minutes of the fixed costs!

    3. As you suspected there must be gone something wrong. Therefore I'll give you the equations which you must solve for x or v respectively:

    $\displaystyle \dfrac{12000}{x^2} - 12 =0$


    $\displaystyle 60v-\dfrac{181200}{v^2} = 0$

    Good luck!
    So I solved for x and got 31.62 for the 1st question.
    And for the second one I have no idea how to solve it.
    My math skills arent the best but I need this paper for my civil engineering diploma. This is a great help.
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  6. #6
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    Quote Originally Posted by stacyg View Post
    So I solved for x and got 31.62 for the 1st question. <<<< OK
    And for the second one I have no idea how to solve it.
    My math skills arent the best but I need this paper for my civil engineering diploma. This is a great help.
    I'll show you how you could solve the 2nd equation:

    $\displaystyle 60v-\dfrac{181200}{v^2} = 0$ Multiply through by v:

    $\displaystyle 60v^3-181200 = 0$ Move the constant to the RHS by adding 181200

    $\displaystyle 60v^3 = 181200$ Divide through by 60:

    $\displaystyle v^3 = 3020$ Now take the cube root on both sides:

    $\displaystyle v = \sqrt[3]{3020} \approx 14.454$

    You now know the optimal speed of the ferry. Plug in this value into C(v) to get the optimal costs.
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