StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 45

Expert-verifiedFound in: Page 108

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find a formula for the cost $C\left(r\right)$ of producing a gourmet soup can with radius $r$ and height $5$ inches, and answer the following questions:

- What is the radius of a can that is $5$ inches tall and costs $30$ cents to produce?
- Your manager wants you to produce $5$-inch-tall cans that cost between $20$ and $40$ cents. Write this requirement as an absolute value inequality.
- What range of radii would satisfy your manager? Write an absolute value inequality whose solution set lies inside this range of radii.

- The radius of a can that is $5$ inches tall and costs $30$ cents to produce is,$1.94$ inches.
- The requirement as an absolute value inequality is, $\left|C\left(r\right)-30\right|<10$.
- The range of radius that is satisfied is, $r\in \left(1.43,2.38\right)$. An absolute value inequality whose solution set lies inside this range of radii is, $\left|r-1.94\right|<0.44$.

Radius $=r$.

Height $=5$ inches.

Total cents is, $30$.

$C\left(r\right)=0.25\left(2\mathrm{\pi rh}\right)+0.50\left(2{\mathrm{\pi r}}^{2}\right)+0.1\left(2\left(2\mathrm{\pi r}\right)+h\right)$

Substitute $h=5$ in the cost function.

$30=0.25\left(2\mathrm{\pi r}\left(5\right)\right)+0.50\left(2{\mathrm{\pi r}}^{2}\right)+0.1\left(2\left(2\mathrm{\pi r}\right)+5\right)\phantom{\rule{0ex}{0ex}}30=2.5\mathrm{\pi r}+0.1{\mathrm{\pi r}}^{2}+0.4\mathrm{\pi r}+0.5\phantom{\rule{0ex}{0ex}}0=2.9\mathrm{\pi r}+0.1{\mathrm{\pi r}}^{2}+0.5-30\phantom{\rule{0ex}{0ex}}0=2.9\mathrm{\pi r}+0.1{\mathrm{\pi r}}^{2}-29.5\phantom{\rule{0ex}{0ex}}\mathrm{r}\approx 1.94$

Hence, the radius is, $1.94$ inches.

The manager wants us to produce $5$-inch-tall cans that cost between $20$ and $40$ cents.

$20<\left|C\right(r\left)\right|<40\phantom{\rule{0ex}{0ex}}20-20<\left|C\right(r)-20|<40-20\phantom{\rule{0ex}{0ex}}0<\left|C\right(r)-20|<20$

Again, subtracting $10$ from each sides we get,

role="math" localid="1648026622152" $\left|C\right(r)-20-10|<20-10\phantom{\rule{0ex}{0ex}}\left|C\right(r)-30|<10$

Hence, the inequality is, $\left|C\left(r\right)-30\right|<10$.

$r\in \left(1.43,2.38\right)$.

Therefore, $\left|r-1.94\right|<0.44$.

94% of StudySmarter users get better grades.

Sign up for free