Question

World renowned ice cream entrepreneurs Sfiso and Richard produce two types of premium

dairy ice cream products, Sfiso n' Richard’s Chocolate Concussion and Vanilla Brain Freeze.

Their chocolate ice cream requires 6𝑙𝑙 milk and 8𝑙𝑙 of peanuts per litre while the vanilla option

requires 9𝑙𝑙 milk and 5𝑙𝑙 peanuts per litre. Sfiso and Richard currently enjoy a surplus of all

other ingredients required for their ice cream but only have 360𝑙𝑙 of milk and 400𝑙𝑙 of peanuts

for this limited production run. Given that the entrepreneurs charge 𝑅𝑅5 for each container of

Chocolate Concussion and 𝑅𝑅7 for each Vanilla Brain Freeze, how many of each type should

Sfiso and Richard produce in order to maximize their profit and what is the maximum?

Approximate your answer to the nearest tenth of a litre.

Answer 1

$\displaystyle \text{Let $x_1$ represent chocolate ice cream and $x_2$ vanilla}\\ \text{The linear program in the given problem is }\\ Maximize: 5x_1 +7x_2\\ \text{Subject to: } 6x_1 +9x_2 \leq 360\\ \qquad \qquad \quad 8x_1 + 5x_2 \leq 400\\\text{The linear program in its standard form is }\\ Maximize: 5x_1 +7x_2+0x_3+0x_4\\ \text{Subject to: } 6x_1 +9x_2 +x_3 = 360\\ \qquad \qquad \quad 8x_1 + 5x_2 + x_4= 400\\ \text{Next we form our first Tableau from our linear program}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 & \\ & & 5 & 7 & 0 & 0 &\\\hline x_3 & 0 & 6 & 9 & 1 & 0 & 360 \\\hline x_4 & 0 & 8 & 5 & 0 & 1 &400\\\hline & & -5 & -7 & 0 & 0 &0\\ \end{matrix}\\ \text{Next, we locate the most negative number in the bottom row(-7), and label the column}\\ \text{where it is found the work column , we then form positive ratios by dividing the elements}\\ \text{in the work column by corresponding elements in the last column. Next we label the }\\ \text{the smallest positive ratio, the pivot element. Using elementary row operations we }\\ \text{reduce the pivot element to 1 and other elements in the work column to 0 to obtain}\\ \text{our tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 & \\ & & 5 & 7 & 0 & 0 &\\\hline x_2 & 7 & \frac{2}{3} & 1 & \frac{1}{9} & 0 & 40 \\\hline x_4 & 0 & \frac{14}{3} & 0 & -\frac{5}{9} & 1 &200\\\hline & & -\frac{1}{3} & 0 & \frac{7}{9} & 0 &280\\ \end{matrix}\\ \text{We observe that there is still a negative element in the bottom row, repeating the process }\\ \text{we obtain tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 & \\ & & 5 & 7 & 0 & 0 &\\\hline x_2 & 7 & 0 & 1 & \frac{4}{21} & -\frac{1}{7} & 11\frac{3}{7} \\\hline x_1 & 5 & 1 & 0 & -\frac{15}{126} & \frac{3}{14} &42\frac{6}{7}\\\hline & & 0 & 0 & \frac{31}{42} & \frac{1}{14} &294\frac{2}{7}\\ \end{matrix}\\ \text{Since there are no negative elements in the bottom row, the solution is feasible. }\\ \text{The company should produce 5 of chocolate ice cream and 7 of vanilla}\\ \text{The maximum values is 290 to the nearest tenth}$