Solving "Save the Pups" with Lagrange Multipliers

According to the method of Lagrange Multipliers, to maximize F(x,y) subject to g(x,y) = C we seek x, y, and λ such that

ΔF = λ·Δg

Here we have M and N instead of x and y, and

F(M,N) = M² + NM + 12N while the constraint is g(M,N)=1000 where g(M,N) = 25M + 5N.

We have three equations:

(Eq-1) F_M = λ· g_M. In this case this equation becomes: 2M + N = λ·25
(Eq-2) F_N = λ· g_N. In this case this equation becomes: M + 12 = λ· 5
(Eq-3) 25M + 5N = 1000.

Using Eq-3 one obtains N = 200 - 5M which, when substituted into Eq-1, yields: 2M + (200-5M) = 25λ

Meanwhile, Eq-2 can be re-written as 5M + 60 = 25λ

Thus we get 200- 3M = 5M + 60 → 200 = 8M + 60 → 140 = 8M → M = 140÷8 = 70÷4 = 35÷2 = 17.5 gallons of medicine, as before. As in the other solutions, we find N = 112.5 gallons or medicine, and F(M,N) = 3,625 lives saved is the maximum.

But why resort to multi-variable calculus methods when simple Calculus I optimization methods work?

Then again, by using the vertex form of a parabola, we can obtain a pre-calculus solution to Save the Pups. Note however that this pre-calculus method works because the function we are trying to maximize is a parabola. For general functions, we will need to use the Calculus I optimization method (or Lagrange, if we want to "shoot flies with cannons")