To formulate the Linear Programming $($LP$)$ problem for Waggin Tails dog food company, we'll need to define the decision variables, the objective function, and the constraints. Here's how to complete the formulation:

### Decision Variables

$1.**$Regular Chow Variables:$**$

$-$ Let $\backslash ($ x$\_1\backslash )$ be the pounds of Regular Chow produced.

$2.**$Premium Chow Variables:$**$

$-$ Let $\backslash ($ x$\_2\backslash )$ be the pounds of Premium Chow produced.

$3.**$Ingredient Variables:$**$

$-$ Let $\backslash ($ a$\_\{$i$1\}\backslash )$ be the pounds of Beef A used in Regular Chow.

$-$ Let $\backslash ($ b$\_\{$i$1\}\backslash )$ be the pounds of Beef B used in Regular Chow.

$-$ Let $\backslash ($ c$\_\{$i$1\}\backslash )$ be the pounds of Beef C used in Regular Chow.

$-$ Let $\backslash ($ d$\_\{$i$1\}\backslash )$ be the pounds of Cereal used in Regular Chow.

$-$ Similarly, let $\backslash ($ a$\_\{$i$2\}\backslash ),\backslash ($ b$\_\{$i$2\}\backslash ),\backslash ($ c$\_\{$i$2\}\backslash ),$ and $\backslash ($ d$\_\{$i$2\}\backslash )$ be the pounds of Beef A$,$ Beef B$,$ Beef C$,$ and Cereal used in Premium Chow.

### Objective Function

The objective is to maximize profit. The profit is calculated as the revenue from selling the chow minus the cost of the ingredients.

Revenue from Regular Chow: $\backslash (10\backslash$times x$\_1\backslash )$

Revenue from Premium Chow: $\backslash (14\backslash$times x$\_2\backslash )$

Let $\backslash ($ C$\_$a$,$ C$\_$b$,$ C$\_$c$,\backslash )$ and $\backslash ($ C$\_$d $\backslash )$ be the cost per pound of Beef A$,$ Beef B$,$ Beef C$,$ and Cereal, respectively.

The cost of ingredients for Regular Chow: $\backslash ($ C$\_$a $\backslash$times a$\_\{$i$1\}+$ C$\_$b $\backslash$times b$\_\{$i$1\}+$ C$\_$c $\backslash$times c$\_\{$i$1\}+$ C$\_$d $\backslash$times d$\_\{$i$1\}\backslash )$

The cost of ingredients for Premium Chow: $\backslash ($ C$\_$a $\backslash$times a$\_\{$i$2\}+$ C$\_$b $\backslash$times b$\_\{$i$2\}+$ C$\_$c $\backslash$times c$\_\{$i$2\}+$ C$\_$d $\backslash$times d$\_\{$i$2\}\backslash )$

Profit function:

$\backslash [$

$\backslash$text$\{$Profit$\}=(10\backslash$times x$\_1+14\backslash$times x$\_2)-\backslash$left$[($C$\_$a $\backslash$times a$\_\{$i$1\}+$ C$\_$b $\backslash$times b$\_\{$i$1\}+$ C$\_$c $\backslash$times c$\_\{$i$1\}+$ C$\_$d $\backslash$times d$\_\{$i$1\})+($C$\_$a $\backslash$times a$\_\{$i$2\}+$ C$\_$b $\backslash$times b$\_\{$i$2\}+$ C$\_$c $\backslash$times c$\_\{$i$2\}+$ C$\_$d $\backslash$times d$\_\{$i$2\})\backslash$right$]$

$\backslash ]$

### Constraints

$1.**$Ingredient Constraints $($Limited Amounts$)$:$**$

$-$ Let $\backslash ($ A$,$ B$,$ C $\backslash )$ be the total available pounds of Beef A$,$ Beef B$,$ and Beef C$,$ respectively.

$**$Beef A constraint:$**$

$\backslash [$

a$\_\{$i$1\}+$ a$\_\{$i$2\}\backslash$leq A

$\backslash ]$

$**$Beef B constraint:$**$

$\backslash [$

b$\_\{$i$1\}+$ b$\_\{$i$2\}\backslash$leq B

$\backslash ]$

$**$Beef C constraint:$**$

$\backslash [$

c$\_\{$i$1\}+$ c$\_\{$i$2\}\backslash$leq C

$\backslash ]$

$2.**$Cereal Constraints:$**$

$-$ Since there is unlimited cereal, no upper limit is needed. However, cereal is restricted by the percentage constraints in the chow formulas.

$3.**$Regular Chow Composition Constraints:$**$

$-$ Regular Chow can be at most $50\%$ cereal by weight. Hence, if $\backslash ($ W$\_\{$R$\}\backslash )$ is the total weight of Regular Chow:

$\backslash [$

d$\_\{$i$1\}\backslash$leq $0\mathrm{.}5\backslash$times x$\_1$

$\backslash ]$

$-$ Regular Chow weight constraint:

$\backslash [$

a$\_\{$i$1\}+$ b$\_\{$i$1\}+$ c$\_\{$i$1\}+$ d$\_\{$i$1\}=$ x$\_1$

$\backslash ]$

$4.**$Premium Chow Composition Constraints:$**$

$-$ Premium Chow can be at most $25\%$ cereal by weight. Hence, if $\backslash ($ W$\_\{$P$\}\backslash )$ is the total weight of Premium Chow:

$\backslash [$

d$\_\{$i$2\}\backslash$leq $0\mathrm{.}25\backslash$times x$\_2$

$\backslash ]$

$-$ Premium Chow weight constraint:

$\backslash [$

a$\_\{$i$2\}+$ b$\_\{$i$2\}+$ c$\_\{$i$2\}+$ d$\_\{$i$2\}=$ x$\_2$

$\backslash ]$

$5.**$Beef Grade Constraints:$**$

$-$ Regular Chow requires an average beef grade of at least $3.$ Assuming $\backslash ($ G$\_$A$,$ G$\_$B$,$ G$\_$C $\backslash )$ are the grades of Beefs A$,$ B$,$ and C:

$\backslash [$

$\backslash$frac$\{$G$\_$A $\backslash$times a$\_\{$i$1\}+$ G$\_$B $\backslash$times b$\_\{$i$1\}+$ G$\_$C $\backslash$times c$\_\{$i$1\}\}\{$a$\_\{$i$1\}+$ b$\_\{$i$1\}+$ c$\_\{$i$1\}\}\backslash$geq $3$

$\backslash ]$

To handle this, you can use a constraint formulation involving:

$\backslash [$

G$\_$A $\backslash$times a$\_\{$i$1\}+$ G$\_$B $\backslash$times b$\_\{$i$1\}+$ G$\_$C $\backslash$times c$\_\{$i$1\}\backslash$geq $3\backslash$times $($a$\_\{$i$1\}+$ b$\_\{$i$1\}+$ c$\_\{$i$1\})$

$\backslash ]$

$-$ Premium Chow requires an average beef grade of at least $4$:

$\backslash [$

$\backslash$frac$\{$G$\_$A $\backslash$times a$\_\{$i$2\}+$ G$\_$B $\backslash$times b$\_\{$i$2\}+$ G$\_$C $\backslash$times c$\_\{$i$2\}\}\{$a$\_\{$i$2\}+$ b$\_\{$i$2\}+$ c$\_\{$i$2\}\}\backslash$geq $4$

$\backslash ]$

Similarly:

$\backslash [$

G$\_$A $\backslash$times a$\_\{$i$2\}+$ G$\_$B $\backslash$times b$\_\{$i$2\}+$ G$\_$C $\backslash$times c$\_\{$i$2\}\backslash$geq $4\backslash$times $($a$\_\{$i$2\}+$ b$\_\{$i$2\}+$ c$\_\{$i$2\})$

$\backslash ]$

### Summary of the LP Formulation

$**$Objective Function:$**$

Maximize:

$\backslash [$

$\backslash$text$\{$Profit$\}=10$x$\_1+14$x$\_2-($C$\_$a $($a$\_\{$i$1\}+$ a$\_\{$i$2\})+$ C$\_$b $($b$\_\{$i$1\}+$ b$\_\{$i$2\})+$ C$\_$c $($c$\_\{$i$1\}+$ c$\_\{$i$2\})+$ C$\_$d $($d$\_\{$i$1\}+$ d$\_\{$i$2\}))$

$\backslash ]$

$**$Constraints:$**$

$1.$ Ingredient limits:

$\backslash [$

a$\_\{$i$1\}+$ a$\_\{$i$2\}\backslash$leq A

$\backslash ]$

$\backslash [$

b$\_\{$i$1\}+$ b$\_\{$i$2\}\backslash$leq B

$\backslash ]$

$\backslash [$

c$\_\{$i$1\}+$ c$\_\{$i$2\}\backslash$leq C

$\backslash ]$

$2.$ Cereal constraints:

$\backslash [$

d$\_\{$i$1\}\backslash$leq $0\mathrm{.}5$ x$\_1$

$\backslash ]$

$\backslash [$

d$\_\{$i$2\}\backslash$leq $0\mathrm{.}25$ x$\_2$

$\backslash ]$

$3.$ Weight constraints:

$\backslash [$

a$\_\{$i$1\}+$ b$\_\{$i$1\}+$ c$\_\{$i$1\}+$ d$\_\{$i$1\}=$ x$\_1$

$\backslash ]$

$\backslash [$

a$\_\{$i$2\}+$ b$\_\{$i$2\}+$ c$\_\{$i