I was trying to find an answer to the following optimization problem, but when testing out values for my solution, I found out that my answer was wrong. I can't find my mistake.
Question:
$T = c(t+2.27) + t$
$N= (33t-15)(33t-14)^c$
Optimize $T$ over the solution set of $N$. Express your answer as a relationship between $t$ and $N$
Solution that I got:
$∇T = ∇N$
$∇[c(t+2.27) + t] = a∇[(33t-15)(33t-14)^c]$
Calculate partial derivatives:
$dT/dc = t+2.27$
$dT/dt = c+1$
$dN/dc = (33t-15)(33t-14)^cln(33t-14)$
$dN/dt = 33c(33t-15)(33t-14)^c/(33t-14) +33(33t-14)^c$
solve for the system of equations:
$t+2.27 = a(33t-15)(33t-14)^cln(33t-14)$
$c+1 = a(33c)(33t-15)(33t-14)^c/(33t-14) +33(33t-14)^c$
$N = (33t-15)(33t-14)^c$
For simplicity's sake, let $33t-14 = g$
$t+2.27 = a(g-1)(g)^cln(g)$
$c+1 = a(33c)(g-1)(g)^c/(33t-14) +33(g)^c$
$N = (g-1)(g)^c$
divide the first 2 equations to eliminate a:
$(t+2.27)/(c+1 - 33g^c) =gln(g)/33c$
$33c(t+2.27) = c(g)ln(g)+gln(g)(1-33g^c)$
$c(33t+74.91-glng) = gln(g)(1-33g^c)$
From our constraint equation, we know that $N = (g-1)(g)^c$
Therefore, $c = log[N/(g-1)]/log(g)$ and $g^c = N/(g-1)$
Substituting that into the equation:
$[log[N/(g-1)]/log(g)](33t+74.91-glng) = gln(g)[1-33N/(g-1)]$
$[log[N/(g-1)]/log(g)](33t+74.91-glng) - gln(g)[1-33N/(g-1)] = 0$
Replace g with 33t-14:
$[log[N/(33t-15)]/log(33t-14)](33t+74.91-(33t-14)ln(33t-14)) - (33t-14)ln(33t-14)[1-33N/(33t-15)] = 0$
I'm not sure where I went wrong. The equation above is supposed to tell me the optimal value of $t$ based on $N$ such that it produces a minimum value of $T$.
If I set $N$ to be 100, you get 170 for $t$, -.466 for $c$, and 90 for $T$.
However, that's not the correct answer because for $N$ = 100, if you set $t$ = 1.5 and $c$ = 0.3, you get a $T$ of 2.624, which is supposed to be the right answer (the lowest value of $T$ possible).
Please let me know what mistake I made while calculating the optimization using Lagrange Multipliers. Thank you!
I'm so dumb, I didn't multiply the whole of the 2nd equation by a, I just multiplied the first term. a is supposed to be the Lagrange multiplier so it should multiply the whole thing.