Question

After taking nutrition class, a big Annie’s Mac and Cheese fan decides to improve the levels of protein and fiber in her favourite lunch by adding broccoli and canned chicken. The nutritional information for the foods referred to in this exercise are given in the table below.

Nutrition Information per serving
Nutrient	Mac and Cheese	Broccoli	Chicken	Shells
Calories Protein (g) Fiber (g)	270 10 2	51 5.4 5.2	70 15 0	260 9 5

1. (M) If she wants to limit her lunch to 400 calories but get 30 g of protein and 10 g of fiber, what properties of servings of Mac and Chees, broccoli, and chicken should she use?
2. (M) She found that there was too much broccoli in the proportions from part (a), so she decided to switch from classical Mac and Cheese to Annie’s Whole Wheat Shells and White Cheddar. What proportions of servings of each food should she use to meet the same goals as in part (a)?

Short answer

1. She should use \(0.99\) servings of Mac and Cheese, \(1.54\) servings of broccoli, and \(0.79\) servings of chicken.
2. Alternatively, she can use \(0.88\) servings of broccoli, \(1.03\) servings of chicken, and \(1.09\) servings of Whole Wheat Shells and White Cheddar.

Step-by-step solution

Write the given data in the matrix form

Let the matrix for Mac and Cheese be MC. Then, \(MC = \left( {\begin{array}{*{20}{c}}{270}\\{10}\\2\end{array}} \right)\).

Let the matrix for broccoli be B. Then, \(B = \left( {\begin{array}{*{20}{c}}{51}\\{5.4}\\{5.2}\end{array}} \right)\).

Let the matrix for chicken be C. Then, \(C = \left( {\begin{array}{*{20}{c}}{70}\\{15}\\0\end{array}} \right)\).

Let the matrix for Whole Wheat Shells and White Cheddar be S. Then, \(S = \left( {\begin{array}{*{20}{c}}{260}\\9\\5\end{array}} \right)\).

Express the system for part (a)

(a)

Let \({x_1},{x_2},\) and \({x_3}\) be the number servings of Mac and Cheese, broccoli, and chicken, respectively.

The matrix expression for this problem is \(MC{x_1} + B{x_2} + C{x_2} = \left( {\begin{array}{*{20}{c}}{400}\\{30}\\{10}\end{array}} \right)\). That is,

\(\left( {\begin{array}{*{20}{c}}{270}\\{10}\\2\end{array}} \right){x_1} + \left( {\begin{array}{*{20}{c}}{51}\\{5.4}\\{5.2}\end{array}} \right){x_2} + \left( {\begin{array}{*{20}{c}}{70}\\{15}\\0\end{array}} \right){x_3} = \left( {\begin{array}{*{20}{c}}{400}\\{30}\\{10}\end{array}} \right)\).

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{270}&{51}&{70}&{400}\\{10}&{5.4}&{15}&{30}\\2&{5.2}&0&{10}\end{array}} \right)\).

Solve the system

First, reduce the augmented matrix into the row echelon form.

Interchange rows one and three, i.e., \({R_1} \leftrightarrow {R_3}\).

\(\left( {\begin{array}{*{20}{c}}{270}&{51}&{70}&{400}\\{10}&{5.4}&{15}&{30}\\2&{5.2}&0&{10}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}2&{5.2}&0&{10}\\{10}&{5.4}&{15}&{30}\\{270}&{51}&{70}&{400}\end{array}} \right)\)

Divide row one by two.

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.6}&0&5\\{10}&{5.4}&{15}&{30}\\{270}&{51}&{70}&{400}\end{array}} \right)\)

At row two, multiply row one by 10 and subtract row two from it, i.e., \({R_2} \to 10{R_1} - {R_2}\). And at row three, multiply row one by 270 and subtract row three from it, i.e., \({R_3} \to 270{R_1} - {R_3}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.6}&0&5\\0&{20.6}&{ - 15}&{20}\\0&{651}&{ - 70}&{950}\end{array}} \right)\)

Divide row two by 20.6.

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.6}&0&5\\0&1&{ - \frac{{75}}{{103}}}&{\frac{{100}}{{103}}}\\0&{651}&{ - 70}&{950}\end{array}} \right)\)

At row three, multiply row two by 651 and subtract row three from it, i.e., \({R_3} \to 651{R_2} - {R_3}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.6}&0&5\\0&1&{ - \frac{{75}}{{103}}}&{\frac{{100}}{{103}}}\\0&0&{ - \frac{{41615}}{{103}}}&{ - \frac{{32750}}{{103}}}\end{array}} \right)\)

This implies,

\(\begin{array}{c} - \frac{{41615}}{{103}}{x_3} = - \frac{{32750}}{{103}}\\{x_3} = \frac{{6550}}{{8323}}\end{array}\)

\(\begin{array}{c}{x_2} - \frac{{75}}{{103}}{x_3} = \frac{{100}}{{103}}\\{x_2} = \frac{{100}}{{103}} + \frac{{75}}{{103}}\left( {\frac{{6550}}{{8323}}} \right)\\{x_2} = \frac{{12850}}{{8323}}\end{array}\)

\(\begin{array}{c}{x_1} + 2.6{x_2} = 5\\{x_1} = 5 - 2.6\left( {\frac{{12850}}{{8323}}} \right)\\{x_1} = \frac{{8205}}{{8323}}\end{array}\)

That is, \({x_1} \approx 0.99,{x_2} \approx 1.54,\) and \({x_3} \approx 0.79\).

Express the system for part (b)

(b)

Let \({x_4}\) be the number of servings of Whole Wheat Shells and White Cheddar.

The matrix expression for this problem is \(B{x_2} + C{x_3} + S{x_4} = \left( {\begin{array}{*{20}{c}}{400}\\{30}\\{10}\end{array}} \right)\). That is,

\(\left( {\begin{array}{*{20}{c}}{51}\\{5.4}\\{5.2}\end{array}} \right){x_2} + \left( {\begin{array}{*{20}{c}}{70}\\{15}\\0\end{array}} \right){x_3} + \left( {\begin{array}{*{20}{c}}{260}\\9\\5\end{array}} \right){x_4} = \left( {\begin{array}{*{20}{c}}{400}\\{30}\\{10}\end{array}} \right)\).

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{51}&{70}&{260}&{400}\\{5.4}&{15}&9&{30}\\{5.2}&0&5&{10}\end{array}} \right)\).

Solve the system

First, reduce the augmented matrix into the row echelon form.

Interchange rows one and three, i.e., \({R_1} \leftrightarrow {R_3}\).

\(\left( {\begin{array}{*{20}{c}}{51}&{70}&{260}&{400}\\{5.4}&{15}&9&{30}\\{5.2}&0&5&{10}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}{5.2}&0&5&{10}\\{5.4}&{15}&9&{30}\\{51}&{70}&{260}&{400}\end{array}} \right)\)

Divide row 1 by 5.2.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{5}{{5.2}}}&{\frac{{10}}{{5.2}}}\\{5.4}&{15}&9&{30}\\{51}&{70}&{260}&{400}\end{array}} \right)\)

At row two, multiply row one by 5.4 and subtract row two from it, i.e., \({R_2} \to 5.4{R_1} - {R_2}\), and at row three, multiply row one by 51 and subtract row three from it, i.e., \({R_3} \to 51{R_1} - {R_3}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{5}{{5.2}}}&{\frac{{10}}{{5.2}}}\\0&{ - 15}&{ - \frac{{99}}{{26}}}&{ - \frac{{255}}{{13}}}\\0&{ - 70}&{ - \frac{{5485}}{{26}}}&{ - \frac{{3925}}{{13}}}\end{array}} \right)\)

Divide row two by 15.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{5}{{5.2}}}&{\frac{{10}}{{5.2}}}\\0&{ - 1}&{ - \frac{{33}}{{130}}}&{ - \frac{{17}}{{13}}}\\0&{ - 70}&{ - \frac{{5485}}{{26}}}&{ - \frac{{3925}}{{13}}}\end{array}} \right)\)

At row three, multiply row two by 70 and subtract row three from it, i.e., \({R_3} \to 70{R_2} - {R_3}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{5}{{5.2}}}&{\frac{{10}}{{5.2}}}\\0&{ - 1}&{ - \frac{{33}}{{130}}}&{ - \frac{{17}}{{13}}}\\0&0&{\frac{{5023}}{{26}}}&{\frac{{2735}}{{13}}}\end{array}} \right)\)

This implies,

\(\begin{array}{c}\frac{{5023}}{{26}}{x_4} = \frac{{2735}}{{13}}\\{x_4} = \frac{{5470}}{{5023}}\end{array}\)

\(\begin{array}{c} - {x_3} - \frac{{33}}{{130}}{x_4} = - \frac{{17}}{{13}}\\{x_3} = \frac{{17}}{{13}} - \frac{{33}}{{130}}\left( {\frac{{5470}}{{5023}}} \right)\\ = \frac{{853910 - 180510}}{{652990}}\\{x_3} = \frac{{5180}}{{5023}}\end{array}\)

\(\begin{array}{c}{x_2} + \frac{5}{{5.2}}{x_4} = \frac{{10}}{{5.2}}\\{x_2} = \frac{{10}}{{5.2}} - \frac{5}{{5.2}}\left( {\frac{{5470}}{{5023}}} \right)\\ = \frac{{50230 - 27350}}{{26119.6}}\\{x_2} = \frac{{4400}}{{5023}}\end{array}\)

That is, \({x_2} \approx 0.88,{x_3} \approx 1.03,\)and \({x_4} \approx 1.09\).