Can you loop through or map 3+ different data sets and find 3+ different nonlinear optimal results using R's ROI package?

I have a simple non-linear example listed below that properly calculates 1 data set perfectly using the ROI package. In the code, I am filtering on CompanyA only. The goal is to minimize total cost by hiring the optimal number of full-time associates for each company separately.

Each company has the ability to use an outside vendor, which entails a higher wage rate. The benefit of the outside vendor is that you are able to adjust hours easier, since full-time associates require 40 hours each week.

The functions below work like a charm for 1 data set, but I am unable to figure out how to loop through 3 different data sets which produces 3 different Minimized Total Costs and 3 different Optimal full-time associates for each Company. The minimal total cost for Company A, B, & C are $9,000, $8,100, and $13,200, respectively. The Optimal number of associates for Company A, B, and C are 2.92, 2.50, and 3.74.

The last part of the code shows listed the desired output. The name of the tibble is called desired_final_tbl.

# 1.0 LIBRARIES ----

# Solver Backend
library(ROI)
library(ROI.plugin.alabama)

# Core
library(tidyverse)

#1.1 Create tibble data set ----

# Company column - 3 separate companies trying to optimize number of Full Time Associates they should hire
# Week - number of weeks we are analyzing
# Full_Time_Associate_Wage_Rate - wage rate for a full time associate that is guarateed 40 hours
# Outside Vendor Wage Rate - wage rate if we decided to hire help from a third party.
# Cleaning Hours - number of cleaning hours needed for each of the days

data_tbl <- tibble(Company = rep(c("CompanyA", "CompanyB", "CompanyC"), times = c(5,5,5)),
                         Week     = rep(1:5, length.out=15),
                         Full_Time_Associate_Wage_Rate = rep(c(12, 11, 12), times = c(5,5,5)),
                         Outside_Vendor_Wage_Rate = rep(c(15, 13, 15), times = c(5,5,5)),
                         Cleaning_Hours = c(100, 120, 140, 160, 180, 100, 120, 140, 160, 180, 140, 160, 200, 240, 280))

data_tbl

# 2.0 FUNCTIONS EO ----

calc_FT_Associates <- function(Optimal_Full_Time_Associates) {

    data_tbl %>%
        filter(Company == "CompanyA") %>%

        select(Company, Week, Full_Time_Associate_Wage_Rate, Outside_Vendor_Wage_Rate, Cleaning_Hours) %>%

        # Calculate FT Associate Hours
        mutate(FT_Associate_Hours = Optimal_Full_Time_Associates * 40) %>%

        # Calculate Number of Hours the Outside Vendor has to fill
        mutate(Outside_Vendor_Hours = case_when(Cleaning_Hours <= FT_Associate_Hours ~ 0,
                                                TRUE ~ Cleaning_Hours - FT_Associate_Hours)) %>%

        # Calculate Cost of FT Associates
        mutate(FT_Associate_Cost = FT_Associate_Hours*Full_Time_Associate_Wage_Rate) %>%

        # Calculate Cost of Outside Vendor
        mutate(Outside_Vendor_Cost = Outside_Vendor_Hours*Outside_Vendor_Wage_Rate) %>%

        # Calculate Total Cost
        mutate(Total_Cost = FT_Associate_Cost + Outside_Vendor_Cost) %>%

        # Summarise Total Cost
        summarise(Total_Cost_sum = sum(Total_Cost)) %>%

        # Pull summarization into a vector so that it will be able to get passed in below
        pull(Total_Cost_sum)

    }

# Test function for Company A only
calc_FT_Associates(2.92)

# 3.0 OBJECTIVE EO ----

n <- 1

single_data_set_optimization <- function(data = data) {

    hrs_model_nlp <- OP(
        objective   = F_objective(F = calc_FT_Associates, n = n, names = "Opt_Associates"),
        constraints = rbind(
            F_constraint(F = calc_FT_Associates, dir = ">=", rhs = 0)
        ),
        maximum = FALSE
    )

    sol <- ROI_solve(hrs_model_nlp, solver = "alabama", start = rep(1/n, n))

    return(
        bind_cols(
            tibble(Total_Cost  = calc_FT_Associates(sol$solution)),
            enframe(sol$solution) %>% spread(key = name, value = value))
    )
}

single_data_set_optimization(Company_A)


# I need the code above to loop through each data set, find the optimal associates for each company and calculate the minimal Total Cost.
# The tibble below is the desired final result

desired_final_tbl <- tibble(Company = c("CompanyA", "CompanyB", "CompanyC"),
                        Total_Cost = c(9000, 8100, 13200),
                        Opt_Associates = c(2.92, 2.5, 3.74))

desired_final_tbl

Desired Output Image: desired output