我不认为optim可以解决你的问题，因为这个问题是一个带有max_budget约束的 * 约束优化问题。一个有希望的候选者是constrOptim。
有两件事你需要注意：
1.因为你已经指定了fnscale=-1，你不需要在你的objective_function中返回-tot_ROI;否则，返回到最小化过程。
1.你应该有一个sum(spend) <= max_budget的显式语句，这在你的约束优化问题中很关键。

代码

下面是一个实现（我想把objective_function从budget_optimal_allocation中移出来，因为如果你需要改变目标函数，这样更容易维护代码）

# Objective function to maximize total ROI
objective_function <- function(spend, a, b, c) {
    # Calculate total revenue and total spend
    revenue <- a / (1 + b * (spend)^c)
    tot_revenue <- sum(revenue)
    tot_spend <- sum(spend)

    # Calculate total ROI
    tot_revenue / tot_spend
}

budget_optimal_allocation <- function(data, max_budget) {
    # Extract channel names and parameters
    channel_names <- data$channel_name
    a <- data$a
    b <- data$b
    c <- data$c
    spend_min <- data$spend_min
    spend_max <- data$spend_max

    # Initial guess for spend values (equal allocation for each channel)
    init <- spend_min + sqrt(.Machine$double.eps)
    # Box constraints for spend
    k <- nrow(data)
    ui <- rbind(diag(k), -diag(k), -rep(1, k))
    ci <- c(spend_min, -spend_max, -max_budget)
    # Optimize spend allocation to maximize total ROI with box constraints
    result <- constrOptim(
        theta = init,
        f = objective_function,
        a = a,
        b = b,
        c = c,
        ui = ui,
        ci = ci,
        method = "Nelder-Mead",
        control = list(fnscale = -1)
    )

    # Combine channel names and optimal spend into a data frame
    return(list(
        optimal_allocation = data.frame(
            channel_name = channel_names,
            optimal_spend = result$par
        ),
        max_tot_ROI = result$value
    ))
}

输出

与你问题中的数据相同

# Example data
example_data <- data.frame(
    channel_name = c("Channel 1", "Channel 2", "Channel 3"),
    a = c(5000, 8000, 6000),
    b = c(0.001, 0.002, 0.0015),
    c = c(-1.05, -1.2, -0.95),
    spend_min = c(100, 200, 150),
    spend_max = c(500, 800, 600)
)

# Max budget for optimization
max_budget <- 1000

# Run the optimization function
optimal_allocation <- budget_optimal_allocation(example_data, max_budget)

我们将看到

> optimal_allocation
$optimal_allocation
  channel_name optimal_spend
1    Channel 1           100
2    Channel 2           200
3    Channel 3           150

$max_tot_ROI
[1] 42.2219

如果你想 * 用尽 * 所有的预算

你可以有以下的变化init <- (spend_min + spend_max) / 2，ui <- rbind(diag(k), -diag(k), rep(1, k))和ci <- c(spend_min, -spend_max, max_budget)，所以，代码是

budget_optimal_allocation <- function(data, max_budget) {
    # Extract channel names and parameters
    channel_names <- data$channel_name
    a <- data$a
    b <- data$b
    c <- data$c
    spend_min <- data$spend_min
    spend_max <- data$spend_max

    # Initial guess for spend values (equal allocation for each channel)
    eps <- sqrt(.Machine$double.eps)
    init <- (spend_min + spend_max) / 2
    # Box constraints for spend
    k <- nrow(data)
    ui <- rbind(diag(k), -diag(k), rep(1, k))
    ci <- c(spend_min, -spend_max, max_budget)
    # Optimize spend allocation to maximize total ROI with box constraints
    result <- constrOptim(
        theta = init,
        f = objective_function,
        a = a,
        b = b,
        c = c,
        ui = ui,
        ci = ci,
        method = "Nelder-Mead",
        control = list(fnscale = -1)
    )

    # Combine channel names and optimal spend into a data frame
    return(list(
        optimal_allocation = data.frame(
            channel_name = channel_names,
            optimal_spend = result$par
        ),
        max_tot_ROI = result$value
    ))
}

你会看到输出

> optimal_allocation
$optimal_allocation
  channel_name optimal_spend
1    Channel 1      252.1112
2    Channel 2      427.5322
3    Channel 3      320.3566

$max_tot_ROI
[1] 18.99994

并检查是否所有预算都用完了

> sum(optimal_allocation$optimal_allocation$optimal_spend)
[1] 1000

我认为你的问题可以用optim来解决，但你必须做一些代数运算，因为optim更适合于无约束优化。它可以使用方法L-BFGS-B和Brent相对地处理边界，但没有其他约束比这更复杂。
您可以使用拉格朗日乘子将简单的约束优化问题转化为无约束优化问题。本质上，定义一个与原始函数相同的函数，加上约束时间常数。

L = function(a,b,c,lambda) {objective_function(a,b,c)-lambda*(max_budget-sum(a,b,c))}

最大化L.
我对拉格朗日乘数有点生疏了，但这应该会在你每次超出预算时惩罚目标。lambda越高，你对超出预算的宽容就越少。