CODEWITHSUNDEEP
revalequationsolverdifferentiation
anjali
anjali

Reputation: 9

How to get value of X in equation, d(M)/d(x) =constant using R?

I am trying to find optimal quantity for which I have to equate differentiation of Total Revenue Equation with Marginal cost. I dont know how to solve for x here. Differentiation works on expression type variable and returns the same, and solve() take numeric equation with only coefficient. I dont want to manually input coefficent.

TR = expression(Quantity * (40- 3*Quantity))
MR = D(TR,"Quantity")
Optimal_Quantity = solve(MR-MC) to get Q

The last line is pseudo code on what I want to achieve, Please help. I can manually enter the values, But wish to make it universal. MC = constant numeric value on RHS

Upvotes: 1

Views: 38

Answers (1)

Rui Barradas
Rui Barradas

Reputation: 76651

I am not completely understanding but if you want to optimize a function, find that functions derivative then find the derivative's zeros.

TR <- expression(Quantity * (40- 3*Quantity))
MR <- D(TR,"Quantity")
class(MR)
#> [1] "call"

dTR <- function(x, const) {
  e <- new.env()
  e$Quantity <- x
  eval(MR, envir = e) - const
}

MC <- 0

u <- uniroot(dTR, interval = c(-10, 10), const = MC)
u
#> $root
#> [1] 6.666667
#> 
#> $f.root
#> [1] 0
#> 
#> $iter
#> [1] 1
#> 
#> $init.it
#> [1] NA
#> 
#> $estim.prec
#> [1] 16.66667

curve(dTR(x, const = MC), from = -10, to = 10)
abline(h = 0)
points(u$root, u$f.root, pch = 16, col = "red")

Created on 2022-11-19 with reprex v2.0.2

Edit

To make the function dTR more general purpose, I have included an argument FUN. Above it would only evaluate MR, it can now evaluate any function passed to it.
The code below plots dTR in a large range of values, from -10 to 100, hoping to catch negative and positive end points. Then, after drawing the horizontal axis, boxes the root between 20 and 30.

dTR <- function(x, FUN, const) {
  e <- new.env()
  e$Quantity <- x
  eval(FUN, envir = e) - const
}

total.revenue <- expression(Quantity * (10- Quantity/5))             
marginal.revenue <- D(total.revenue, "Quantity")                                    
marginal.cost <- 1            

curve(dTR(x, FUN = marginal.revenue, const = marginal.cost), from = -10, to = 100)
abline(h = 0)
abline(v = c(20, 30), lty = "dashed")



u <- uniroot(dTR, interval = c(20, 30), FUN = marginal.revenue, const = marginal.cost)
u
#> $root
#> [1] 22.5
#> 
#> $f.root
#> [1] 0
#> 
#> $iter
#> [1] 1
#> 
#> $init.it
#> [1] NA
#> 
#> $estim.prec
#> [1] 7.5

Created on 2022-11-22 with reprex v2.0.2

Upvotes: 0

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