# Thread: Optimization and L'Hospital's rule

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

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

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

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.)

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

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?

2. Originally Posted by Open that Hampster!
1) $y = \lim_{x\to\infty} x^{\frac{ln(7)}{1+ln(x)}}$
$ln(y) = \lim_{x\to\infty} \frac{ln(7)ln(x)}{1+lnx}$
Deriving the right side gives me:
$ln(y) = \lim_{x\to\infty} \frac{\frac{ln(x)}{7}+\frac{ln(7)}{x}}{\frac{1}{x} }$ $=> \frac{\infty + 0}{0}$

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

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.)

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

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

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

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

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