My approach is at the bottom
The table at the bottom gives a composition of various foods used in making cereals. The other material in each food is fiber, water, etc. The company blends these food materials and makes two kinds of cereals. In the process of blending, 3% of protein, 5% of starch, and 10% of minerals and vitamins are completely lost from the mix. For each 100 kg of foods added in the blend, the blending process adds 5 kg of other material (mainly water and fat).
Cereal type 1 sells for \$1.50/kg. It should contain at least 22% protein, 2% of minerals and vitamins, and at most 30% of starch by weight. Cereal type 2 sells for \$1.00/kg. It should contain at least 30% starch by weight. The company would like to determine the most profitable product mix that meets the requirements described above.
What I tried: Let $x_1=$ food 1, $x_2=$ food 2, $x_3=$ food 3, $x_4=$ food 4 For cereal 1:
minimize $0.68x_1+0.27x_2+0.31x_3+0.45x_4$
subject to $45x_1+7x_2+12x_3+27x_4-3\geq22$ (protein)
$12x_1+38x_2+25x_3+40x_4-5\geq 30$ (starch)
$4x_1+x_2+2x_3+3x_4-10\geq 2$ (minerals)
For cereal 2:
minimize $0.68x_1+0.27x_2+0.31x_3+0.45x_4$
subject to $12x_1+38x_2+25x_3+40x_4-5\geq 30$ (starch)
$x_1\leq 1500, x_2\leq 500, x_3\leq 1000, x_4\leq 2000$
Then overall, we want to maximize $1.5C_1+C_2$
I don't think that I am doing this correctly but I can't see any other way to do it.
$$\bf\text{Review of your solution}$$ Looking at the objective functions, it seems $x_i$ represents the weight of different foods in $kg$ used to make the cereals per day.
You can't use $x_i$ to denote the amount of food $i$ in both cereal $1$ and cereal $2$ because both cereals have different nutritional requirements and you might obtain an optimum at different values of $x_i$ for both cereals. Use $x_i$ to denote the amount of food $i$ in cereal $1$ and $y_i$ to denote the corresponding amount for food $2$.
Your existing constraints are incorrect as you disregarded the percentages. For example, in the first constraint, the coefficients should be $0.45,0.07,...$. You subtracted $3$ from the amount of protein. Note that the question mentions $3\color{red}\%$ of the protein is lost and the remaining protein should be at-least $22\color{red}\%$ of the total weight of the cereal.
You didn't take into account the availability of different foods per day.
I am not sure what $C_{1,2}$ represent. But according to me, the one and only objective of this linear program should be to maximize the profit of the company assuming all cereal is sold.
$$\bf\text{My solution}$$
The amount of protein added into cereal $1$ is $0.45 x_1+0.7x_2+0.12x_3+0.27x_4$ out of which $3\%$ is lost, so protein present in cereal $1$ is $p_1=0.97[0.45 x_1+0.7x_2+0.12x_3+0.27x_4]$.
Similarly starch and minerals present in cereal $1$ are $s_1=0.95[0.12x_1+0.38x_2+0.25x_3+0.40x_4]$ and $m_1=0.90[0.04x_1+0.01x_2+0.02x_3+0.03x_4]$.
The combined weight of the foods added to the blend is $x_1+x_2+x_3+x_4$ and to this water and fat are added to make the total weight $w_1=1.05[x_1+x_2+x_3+x_4]$.
The cost of making the cereal is $c_1=0.68x_1+0.27x_2+0.31x_3+0.45x_4$ and the income from total sale is $1.50w_1$ so the profit from sale is $P_1=1.5w_1-c_1$.
The constraint on the amount of protein is thus $p_1\ge0.22w_1$. Similarly $s_1\le0.30w_1$ and $m_1\ge0.02w_1$.
For cereal $2$, we only have a constraint on the amount of starch: $s_2\ge0.30w_2$. The profit from sale of cereal $2$ is $P_2=w_2-c_2$. Thus we have the following linear program:
\begin{align*} \text{maximize}&~~\mathbf{\text{Profit} P}=P_1+P_2\\ \text{subject to}&~~\bf\text{Cereal 1 nutritional constraint(s)}\\ &~~\color{blue}{p_1\ge0.22w_1}\\ &~~\color{blue}{s_1\le0.30w_1}\\ &~~\color{blue}{m_1\ge0.03w_1}\\\\ &~~\bf\text{Cereal 2 nutritional constraint(s)}\\ &~~\color{green}{s_2\ge0.30w_2}\\\\ &~~\bf\text{Food availability constraint(s)}\\ &~~\color{red}{x_1+y_1\le1500}\\ &~~\color{red}{x_2+y_2\le500}\\ &~~\color{red}{x_3+y_3\le1000}\\ &~~\color{red}{x_4+y_4\le2000}\\\\ &~~\bf\text{Non-negative constraint(s)}\\ &~~x_i,y_i\ge0\\\\ \text{where}&~~p_1=0.97[0.45 x_1+0.7x_2+0.12x_3+0.27x_4]\\ &~~s_1=0.95[0.12x_1+0.38x_2+0.25x_3+0.40x_4]\\ &~~m_1=0.90[0.04x_1+0.01x_2+0.02x_3+0.03x_4]\\ &~~w_1=1.05[x_1+x_2+x_3+x_4]\\ &~~c_1=0.68x_1+0.27x_2+0.31x_3+0.45x_4\\ &~~P_1=1.5w_1-c_1\\\\ &~~s_2=0.95[0.12y_1+0.38y_2+0.25y_3+0.40y_4]\\ &~~w_2=1.05[y_1+y_2+y_3+y_4]\\ &~~c_2=0.68y_1+0.27y_2+0.31y_3+0.45y_4\\ &~~P_2=w_2-c_2 \end{align*}
Now it is your job to bring all the constraints and objective function in terms of $x_i$ and $y_i$ only.