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Thread: Optimization and L'Hospital's rule

  1. #1
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    Optimization and L'Hospital's rule

    1) $\displaystyle y = \lim_{x\to\infty} x^{\frac{ln(7)}{1+ln(x)}}$
    $\displaystyle ln(y) = \lim_{x\to\infty} \frac{ln(7)ln(x)}{1+lnx}$
    Deriving the right side gives me:
    $\displaystyle ln(y) = \lim_{x\to\infty} \frac{\frac{ln(x)}{7}+\frac{ln(7)}{x}}{\frac{1}{x} }$ $\displaystyle => \frac{\infty + 0}{0}$

    Since the top goes towards a positive number (infinity) and the bottom goes to 0, $\displaystyle ln(y) = \infty => y = e^{\infty}$

    Is my logic wrong?

    2) A cone-shaped paper drinking cup is to be made to hold 33 cm^3 of water. Find the height and radius of the cup that will use the smallest amount of paper. (Give your answers correct to two decimal places.)

    $\displaystyle V = 33 = \frac{1}{3} \pi r^2h$
    $\displaystyle SA = \pi rs + \pi r^2$ where $\displaystyle s = \sqrt{h^2+r^2}$
    So
    $\displaystyle SA = \pi r\sqrt{h^2+r^2}+ \pi r^2$
    $\displaystyle h = \frac{99}{\pi r^2}$
    $\displaystyle => SA = \pi r \sqrt{(\frac{99}{\pi r^2})^2 + r^2}+ \pi r^2$

    Am I starting this off the right way? Because if I am, this looks disgusting. I was trying to derive the Surface Area formula just in terms of r, and then finding the minimum I could plug it back into the volume formula to gind the minimum h?

    Am I wrong in my approach?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Open that Hampster! View Post
    1) $\displaystyle y = \lim_{x\to\infty} x^{\frac{ln(7)}{1+ln(x)}}$
    $\displaystyle ln(y) = \lim_{x\to\infty} \frac{ln(7)ln(x)}{1+lnx}$
    Deriving the right side gives me:
    $\displaystyle ln(y) = \lim_{x\to\infty} \frac{\frac{ln(x)}{7}+\frac{ln(7)}{x}}{\frac{1}{x} }$ $\displaystyle => \frac{\infty + 0}{0}$

    Since the top goes towards a positive number (infinity) and the bottom goes to 0, $\displaystyle ln(y) = \infty => y = e^{\infty}$

    Is my logic wrong?

    2) A cone-shaped paper drinking cup is to be made to hold 33 cm^3 of water. Find the height and radius of the cup that will use the smallest amount of paper. (Give your answers correct to two decimal places.)

    $\displaystyle V = 33 = \frac{1}{3} \pi r^2h$
    $\displaystyle SA = \pi rs + \pi r^2$ where $\displaystyle s = \sqrt{h^2+r^2}$
    So
    $\displaystyle SA = \pi r\sqrt{h^2+r^2}+ \pi r^2$
    $\displaystyle h = \frac{99}{\pi r^2}$
    $\displaystyle => SA = \pi r \sqrt{(\frac{99}{\pi r^2})^2 + r^2}+ \pi r^2$

    Am I starting this off the right way? Because if I am, this looks disgusting. I was trying to derive the Surface Area formula just in terms of r, and then finding the minimum I could plug it back into the volume formula to gind the minimum h?

    Am I wrong in my approach?
    You don't need L.H

    $\displaystyle \ln(y) = \lim_{x\to\infty} \frac{\ln(7) \ln(x)}{1+\ln(x)}= \lim_{x\to\infty} \frac{\ln(7)}{1+\frac{1}{\ln(x)}}= \ln(7)$


    Your start seems fine on the cone except I don't think you need the

    $\displaystyle \pi r^2$ term this would put a top on the cone cup.
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