Question

Imagine you are trying to maximize the calories you burn in a 60 minute workout you do a few times a week. Running burns 9 calories per minute, aerobics burn 6 calories per minute and rowing burns 7 calories per minute. You want to perform all threeexercises to work different muscle groups. For the best effect, you need to run for at least 5 minutes and row for at least 15 minutes.Your aerobics session should be no more than 30 minutes. How many minutes should you perform each exercise to burn themaximum calories?Part A- Write a system of inequalities based on the given constraints. Then, note the objective function. Part B- convert the inequalities into equations, and then use the substitution method to find four possible vertices in the form (x,y,z). Part C- test the four possible vertices that you found in the objective function in part A. Use those values to determine which set of values maximizes the objective function. Show your work and final answer in the space provided.

Answer 1

The problem already have a pathway to guide you to the answer, then let's follow it.

The first part, consists on writing the constrains and the objective function.

We have the following constrains:

- You need to run for at least 5 minutes;

- Row for at least 15 minutes;

- Your aerobics session should be no more than 30 minutes;

-You want to perform all three;

-The total workout is 60 minutes long.

Let's call the running time 'x', the rowing time 'y', and the time associated with the aerobics session 'z'.

Writing those constrains as inequalities goes like this

`\begin{gathered} x\ge5, \\ y\ge15, \\ 1\leq z\leq30, \\ x+y+z=60 \end{gathered}`

And the objective function(what we want to know), is the function that gives us the total amount of calories burnt in 60 minutes.

This function will be the sum of the products of each time by its respective rate.

`f(x,y,z)=9x+6y+7z`

The part B, consists in converting the into equations, and then use the substitution method to find four possible vertices.

We have four possibilites. Fixating 'z' on its minimum value and 'y' on its minimum, we get the following:

`\begin{gathered} z=1\Rightarrow x+y=59 \\ y=15\Rightarrow x=44 \\ (x,y,z)=(44,15,1) \end{gathered}`

Fixating 'z' on its minimum value and 'x' on its minimum, we get the following:

`\begin{gathered} z=1\Rightarrow x+y=59 \\ x=5\Rightarrow y=54 \\ (x,y,z)=(5,54,1) \end{gathered}`

Fixating 'z' on its maximum value and 'y' on its minimum, we get the following:

`\begin{gathered} z=30\Rightarrow x+y=30 \\ y=15\Rightarrow x=15 \\ (x,y,z)=(15,15,30) \end{gathered}`

Fixating 'z' on its maximum value and 'x' on its minimum, we get the following:

`\begin{gathered} z=30\Rightarrow x+y=30 \\ x=5\Rightarrow y=25 \\ (x,y,z)=(5,25,30) \end{gathered}`

On the last part, you just need to test the points we got on our objective function.

The function and the values:

`\begin{gathered} f(x,y,z)=9x+6y+7z \\ \lbrace(44,15,1),(5,54,1),(15,15,30),(5,25,30)\rbrace \end{gathered}`

Calculating the function on each point:

`\begin{gathered} f(44,15,1)=9*44+6*15+7*1=493 \\ f(5,54,1)=376 \\ f(15,15,30)=435 \\ f(5,25,30)=405 \end{gathered}`

Comparing the results, we have the point (44, 15, 1 ) as the one that maximizes the function.