Chapter 12.4, Problem 56SR

a.

To determine

To find: the value of $t$ for an altitude of $4500\text{feet}$ .

Answer to Problem 56SR

The value of $t$ is $31.9\text{ minutes}$ .

Explanation of Solution

Given information:

Given data:

  $t=\frac{40\left(25+1.85a\right)}{50-1.85a}$ and $a=4500\text{feet}$

To calculate the value of $t$ for an altitude of $4500\text{feet}$ ,

  $\begin{array}{l}t=\frac{40\left(25+1.85a\right)}{50-1.85a}\\ \text{Substitute }a=4.5\text{ for }4500\text{ feet then, evaluate}\\ t=40\frac{\left\{25+1.85\left(4.5\right)\right\}}{50-1.85\left(4.5\right)}\\ t=\frac{40\left(25+8.3\right)}{50-8.3}\\ t=\frac{40\left(33.3\right)}{41.7}\\ t=\frac{1332}{41.7}\\ t\approx 31.9\text{ minutes}\end{array}$

Hence, the value of $t$ is $31.9\text{ minutes}$ .

b.

To determine

To calculate: the time to cook a potato at $3500\text{feet}$ for an altitude and at $7000\text{feet}$ and compare the cooking time.

Answer to Problem 56SR

The value of $t$ at $3500\text{feet}$ is $29\text{ minutes}$ .

The value of $t$ at $7000\text{feet}$ is $\text{41}\text{.1 minutes}$ .

The time to cook a potato at $3500\text{feet}$ is less than the time to cook a potato at $7000\text{feet}$ .

Explanation of Solution

Given information:

Given data:

  $t=\frac{40\left(25+1.85a\right)}{50-1.85a}$ , $a_1=3500\text{feet}$ and $a_2=7000\text{feet}$ .

To calculate the value of $t$ for an altitude of $3500\text{feet}$ ,

  $\begin{array}{l}t=\frac{40\left(25+1.85a\right)}{50-1.85a}\\ \text{Substitute }a=3.5\text{ for 3}500\text{ feet then, evaluate}\\ t=40\frac{\left\{25+1.85\left(3.5\right)\right\}}{50-1.85\left(3.5\right)}\\ t=\frac{40\left(25+6.5\right)}{50-6.5}\\ t=\frac{40\left(31.5\right)}{43.5}\\ t=\frac{1260}{43.5}\\ t\approx 29\text{ minutes}\end{array}$

Hence, the value of $t$ is $29\text{ minutes}$ .

Again,

To calculate the value of $t$ for an altitude of $7000\text{feet}$ ,

  $\begin{array}{l}t=\frac{40\left(25+1.85a\right)}{50-1.85a}\\ \text{Substitute }a=7\text{ for 70}00\text{ feet then, evaluate}\\ t=40\frac{\left\{25+1.85\left(7\right)\right\}}{50-1.85\left(7\right)}\\ t=\frac{40\left(25+13\right)}{50-13}\\ t=\frac{40\left(38\right)}{37}\\ t=\frac{1520}{37}\\ t\approx 41.1\text{ minutes}\end{array}$

Hence, the value of $t$ is $\text{41}\text{.1 minutes}$ .

Therefore, the time to cook a potato at $3500\text{feet}$ is less than the time to cook a potato at $7000\text{feet}$ .