Chapter 12.4, Problem 26PPSE

Solution

To find: The time for which each bicycle speed to burn 600 calories per hour.

Bicycling at 11 mph for 22 minutes and at 15 mph for 38 minutes burns 600 calories per hour.

Given information:

Given, a classmate is starting a new fitness program. He is planning to ride his bicycle 60 minutes every day. He burns 7 calories per minute bicycling at 11 mph and 11.75 calories per minute bicycling at 15 mph.

Calculation:

Let he bicycle at 11 mph for x minutes and at 15 mph for y minutes.

Total time is 1 hour i.e., 60 minutes. Hence, $x+y=60$

The calories burnt are given by the equation, $7x+11.75y=600$

This is a system of equations with two equations.

The matrix equation is given by:

  $\left[\begin{array}{cc}1& 1\\ 7& 11.75\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}60\\ 600\end{array}\right]$

Multiplying both sides by the inverse of the matrix on the left gives the required solution.

The determinant of the matrix is given by:

  $\begin{array}{l}D=\left(1\right)\left(11.75\right)-\left(1\right)\left(7\right)\\ D=11.75-7\\ D=4.75\end{array}$

Since the determinant is not 0, the inverse exists.

The inverse of a 2X2 matrix $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $\frac{1}{D}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Plugging the values, the inverse matrix is given by:

  $\frac{1}{4.75}\left[\begin{array}{cc}11.75& -1\\ -7& 1\end{array}\right]=\left[\begin{array}{cc}\frac{11.75}{4.75}& -\frac{1}{4.75}\\ -\frac{7}{4.75}& \frac{1}{4.75}\end{array}\right]$

Multiplying both sides of the equation by the inverse:

  $\begin{array}{l}\left[\begin{array}{cc}\frac{11.75}{4.75}& -\frac{1}{4.75}\\ -\frac{7}{4.75}& \frac{1}{4.75}\end{array}\right]\left[\begin{array}{cc}1& 1\\ 7& 11.75\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\frac{11.75}{4.75}& -\frac{1}{4.75}\\ -\frac{7}{4.75}& \frac{1}{4.75}\end{array}\right]\left[\begin{array}{c}60\\ 600\end{array}\right]\\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\left(\frac{11.75}{4.75}\right)\left(60\right)+\left(-\frac{1}{4.75}\right)\left(600\right)\\ \left(-\frac{7}{4.75}\right)\left(60\right)+\left(\frac{1}{4.75}\right)\left(600\right)\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{705-600}{4.75}\\ \frac{-420+600}{4.75}\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{105}{4.75}\\ \frac{180}{4.75}\end{array}\right]\end{array}$

Hence, $x=\frac{105}{4.75}\approx 22\text{ minutes},y=\frac{180}{4.75}\approx 38\text{ minutes}$

Conclusion:

Bicycling at 11 mph for 22 minutes and at 15 mph for 38 minutes burns 600 calories per hour.