Question #260967

(1) The table below shows the calories, fat, and carbohydrates per ounce for three brands of cereal, and the total amount of calories, fat, and carbohydrates required for a special diet. Calories Fat Carbohydrates Cereal Brand X 50 0 22 Cereal Brand Y 108 0.1 25.5 Cereal Brand Z 127 5.5 18 Total Required 393 5.7 91 (i) Using the information in the table above, write a system of three equations. (ii) Write a system in matrix form Ax b  . (ii) Using the Cramer's Rule or Inverse Method, find the amount of each brand of cereal that will give the level of nutrition required. 


1
Expert's answer
2021-11-11T08:53:04-0500

Using cramer's Rule;


(5010812700.15.52225.518)(xyz)=(3935.791)\begin{pmatrix} 50&108& 127\\ 0&0.1 & 5.5\\ 22&25.5&18 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 393 \\ 5.7\\ 91 \end{pmatrix}



Let D=D= (5010812700.15.52225.518)\begin{pmatrix} 50&108 & 127 \\ 0 & 0.1&5.5\\ 22&25.5&18 \end{pmatrix}



Dx=(3931081275.70.15.59125.518)D_x=\begin{pmatrix} 393& 108&127\\ 5.7&0.1 & 5.5\\ 91&25.5&18 \end{pmatrix}



Dy=(5039312705.75.5229118)D_y=\begin{pmatrix} 50&393 & 127 \\ 0 & 5.7&5.5\\ 22&91&18 \end{pmatrix}



D2=(5010839300.15.72225.591)D_2= \begin{pmatrix} 50&108 & 393 \\ 0&0.1 & 5.7\\ 22&25.5&91 \end{pmatrix}


Det D=500.15.525.51810805.52218+12700.12225.5D=50\begin{vmatrix} 0.1 & 5.5 \\ 25.5 & 18 \end{vmatrix}-108\begin{vmatrix} 0& 5.5 \\ 22 & 18 \end{vmatrix}+127\begin{vmatrix} 0 & 0.1\\ 22 & 25.5 \end{vmatrix}



=50(138.45)108(121)+127(2.2)=50(-138.45)-108(-121)+127(-2.2)

=5866.1=5866.1



Det Dx=3930.15.525.5181085.75.59118+1275.70.19125.5D_x=393\begin{vmatrix} 0.1& 5.5 \\ 25.5& 18 \end{vmatrix}-108\begin{vmatrix} 5.7& 5.5 \\ 91& 18 \end{vmatrix}+127\begin{vmatrix} 5.7 & 0.1\\ 91 & 25.5 \end{vmatrix}


=393(138.45)108(397.9)+127(136.25)=393(-138.45)-108(-397.9)+127(136.25)

=5866.1=5866.1



Det Dy=505.75.5911839305.52218+12705.72291D_y=50\begin{vmatrix} 5.7 & 5.5 \\ 91& 18 \end{vmatrix}-393\begin{vmatrix} 0& 5.5 \\ 22& 18 \end{vmatrix}+127\begin{vmatrix} 0& 5.7 \\ 22 & 91 \end{vmatrix}


=50(397.9)393(121)+127(125.4)=50(-397.9)-393(-121)+127(-125.4)

=11732.2=11732.2



Det D2=500.15.725.59110805.72291+39300.12225.5D_2=50\begin{vmatrix} 0.1 & 5.7 \\ 25.5 & 91 \end{vmatrix}-108\begin{vmatrix} 0& 5.7 \\ 22 & 91 \end{vmatrix}+393\begin{vmatrix} 0& 0.1\\ 22& 25.5 \end{vmatrix}


=50(136.25)108(125.4)+393(22)=50(-136.25)-108(-125.4)+393(-22)

=5866.1=5866.1




x=DxD=5866.15866.1=1,x=\frac{|D_x|}{|D|}=\frac{5866.1}{5866.1}=1,


y=DyD=11732.25866.1=2y=\frac{|D_y|}{|D|}=\frac{11732.2}{5866.1}=2

z=DzD=5866.15866.1=1z=\frac{|D_z|}{|D|}=\frac{5866.1}{5866.1}=1



Amount of each brand of cereal that will give the level of nutrition required


Cereal Brand X=1X=1 ounce

Cereal Brand Y=2Y=2 ounces

Cereal Brand Z=1Z=1 ounce


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