Chapter 12.8, Problem 5P

To determine

To calculate: The cost of running an appliances varies jointly as the number of watts used, the hours of operation,

A $6000$ watt convection oven operates for $10$ min for $15\$$ at a cost of $7.5\$$ per kilowatt-hour. To calculate the cost of cooking two meals if each meal take; $70$ min to cook?

Answer to Problem 5P

Thus, the cost is: $21\text{dollers}$ .

Explanation of Solution

Given information:

The cost of running an appliances varies jointly as the number of watts used, the hours of operation,

A $6000$ watt convection oven operates for $10$ min for $15$ at a cost of $7.5$ per kilowatt-hour. To calculate the cost of cooking two meals if each meal take; $70$ min to cook?

Formula used:

For varies jointly:

  $C=KWhc$

Calculation:

The cost of running an appliances varies jointly as the number of watts used, the hours of operation,

A $6000$ watt convection oven operates for $10$ min for $15$ at a cost of $7.5$ per kilowatt-hour. To calculate the cost of cooking two meals if each meal take; $70$ min to cook?

Recall that for varies jointly:

  $\begin{array}{l}C=KWhc\\ K=\frac{C}{whc}\end{array}$

The values are:

  $\begin{array}{l}C=15,w=6000,\\ h=\frac{10}{16}=\frac{1}{6}\\ c=7.5\end{array}$

Therefore, the given area by,

  $\begin{array}{c}K=\frac{15}{6000\times \frac{1}{6}\times 7.5}\\ K=\frac{15}{1000\times 7.5}\\ K=\frac{15\times 10}{1000\times 75}\\ k=\frac{1}{500}\end{array}$

Then the height is:

  $\begin{array}{c}K=\frac{1}{500}\\ K=0.002\end{array}$

Thus, $C=0.002\left(whc\right)$ .

Hence, when:

  $\begin{array}{c}h=70\times 2\\ =140\\ \frac{140}{6}=\frac{70}{3}\end{array}$

So,

  $\begin{array}{l}C=0.002\left(6000\times \frac{70}{3}\times 7.5\right)\\ C=2100\text{cents}\\ \text{C=21}\end{array}$

Thus, the cost is: $21\text{dollers}$ .