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Thread: Practice problems help:

  1. #1
    Nov 2013

    Practice problems help:

    The top and bottom margins of a poster are 8 cm and the side margins are each 9 cm. If the area of printed material on the poster is fixed at 384 square centimeters, find the dimensions of the poster with the smallest area.

    I did these steps:

    xy = 384
    y = 384/x

    A = (x + 16) * (y + 18)
    A = (x + 16) * (384/x + 18)

    And I appear to be stuck at this step, the book doesn't have an examples listed for this problem either.

    What would the answer be?


    Consider the function whose second derivative is . If and , what is ?

    I've always had trouble with second derivatives, so an answer to this problem with an explanation would greatly help me! Much thanks again!
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  2. #2
    MHF Contributor
    Nov 2010

    Re: Practice problems help:

    #1 $\displaystyle \dfrac{384}{x} = 384x^{-1}$. You will eventually want to take a derivative, and it is much easier to avoid mistakes with the power rule than with the quotient rule.

    $\displaystyle A(x) = (x+16)(384x^{-1}+18)$

    Next, multiply it out. Do you remember the FOIL method? It tells you what to multiply. First Outside Inside Last (FOIL) means multiply the first terms: $\displaystyle x\cdot 384x^{-1}$, then the outside terms: $\displaystyle x\cdot 18$ then the inside terms: $\displaystyle 16\cdot 384x^{-1}$ then the last terms: $\displaystyle 16\cdot 18$. Add them all together. Then take the derivative and set it equal to zero. Next, take the second derivative and plug in any critical values you found. If the second derivative is positive at that value, you found a local minimum.

    #2 I assume you are practicing integration. $\displaystyle \int f''(x)dx = f'(x) + C_1$ where $\displaystyle C_1$ is an arbitrary constant. Since you know $\displaystyle f'(0) = 3$, you can solve for $\displaystyle C_1$ after you integrate by plugging in $\displaystyle 0$ for $\displaystyle x$. Then, you integrate again: $\displaystyle \int f'(x)dx = f(x) + C_2$. You are given $\displaystyle f(0)=4$, so you can solve for $\displaystyle C_2$ after this second integration by plugging in $\displaystyle 0$ for $\displaystyle x$.

    Note: If $\displaystyle \int f''(x)dx = f'(x)+C_1$ then $\displaystyle f'(x) = \int f''(x)dx - C_1$.
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