Q. 45 - Calculus

Question

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:

1.  What is the radius of a can that is $5$ inches tall and costs $30$ cents to produce?
2. 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.
3. What range of radii would satisfy your manager? Write an absolute value inequality whose solution set lies inside this range of radii.

Step-by-Step Solution

Answer

1. The radius of a can that is $5$ inches tall and costs $30$ cents to produce is,$1.94$ inches.
2. The requirement as an absolute value inequality is, $\left|C\left(r\right)-30\right|<10$.
3. 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, $\left|r-1.94\right|<0.44$.

Part a Step 1 . Given information

Radius $=r$.

Height $=5$ inches.

Total cents is, $30$.

Part a Step 2 . Considering the radius r and height 5 inches, the cost function will be,

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

Substitute $h=5$ in the cost function.

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

Hence, the radius is, $1.94$ inches.

Part b Step 1 . Given information

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

Part b Step 2 . Considering that the cost of cans of 5 -inch tall between 20 and 40 cents, the inequality can be written as,

$$\begin{gathered} 20<\left|C\right(r\left)\right|<40 \\ 20-20<\left|C\right(r)-20|<40-20 \\ 0<\left|C\right(r)-20|<20 \end{gathered}$$

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

$$\begin{gathered} \left|C\right(r)-20-10|<20-10 \\ \left|C\right(r)-30|<10 \end{gathered}$$

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

Part c Step 1 . The range of radius that is satisfied is,

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

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