#Multivariable calc help needed asap plz
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the gradient P(s,t) = (6000e^t - 4000s, 6000se^t)
This is the system of equations I have set up that I need to solve:
6000e^t - 4000s = 2 * lambda * s
6000se^t = lambda*e^t
s^2 + e^t = 11/3
I have no idea how to proceed with solving this after thinking about this for a long time...
Please can someone provide the answer with a detailed explanation...
@grave tide u mind taking a look at this when ur awake?
I assume P(s,t)=6000se^t - 2000s^2 + C
L(s,t,λ)=P(s,t)+λ(s^2+e^t-11/3)
∂L/∂s = 6000e^t - 4000s + 2λ = 0
∂L/∂t = 6000se^t + λe^t = 0
∂L/∂λ = s^2 + e^t - 11/3 = 0
then you resolve your values, I got t=ln(8/3), s=1, λ=-6000 as the only valid sol
@raw igloo
Where did u get assume P(s,t) = 6000se^t - 2000s^2 + C from?
the original function
you said your gradient was (6000e^t -4000s, 6000se^t) so I derived the original function from that
@raw igloo
idk whether you had a constant in the original function, but that doesn't affect the Lagrange multipliers
cool
That was the original function
But why r we using the original for lagrange here because they only care about it perpendicular to the boundary
apex predator
Which is just this:
I didn't read the question 💀
They only care about gradient P = lambda*gradientg
I got this system, idk if it's right...
I should consider reading the question at some point
yeah your system looks fine
you can divide out e^t in the second part as e^t≠0
I did this, but this seems very suspect because my numbers I got in the end r so weird lmao
U mind checking if what I did is right?
6000s=lambda
6000e^t - 4000s = 12000s² and e^t=11/3 - s²
22000-6000s²-4000s=12000s²
Where did 22000 come from?
6000(11/3)
then you can use the 6000s=lambda again
getting an egregious cubic
0=72000s³+6s²+4s-22
How do u have 12000s^2* lambda tho? Because we have 2 * lambda * s on the right side
So it should just be 2*(6000s)*s which is just 12000s^2 no?
So there should be no cubic cuz we just have, 22000 - 6000s^2 - 4000s = 12000s^2
Yeah
And then we just apply quadratic formula
9s²+2s-11=0
then you can use t=ln(11/3-s²)
18000s²+4000s-22000=0
18s²+4s-22=0
9s²+2s-11=0
Wow
And that gives x = 1 and x = -11/9
But we ingore the negative
Since the question said only care about s,t >= 0
yes
rumour has it: if your Lagrange multipliers exist on your boundaries then IT'S THE SAME AS BEING PERPENDICULAR!!!!!!
anyways, I was waiting to share that
learning experience
Bro, u are actually a legend. Like actually next lvl talent
I was on the other math discord and not 1 out of the 70,000 people
Knew multivariable calc
Legit no one
lmao
They were all Indians swearing in their language and saying that I'm asking bs questions 😂
lmao
Ok so now that I have that one point of s = 1 and t = Ln (8/3). They asked me to calculate the profit at that point, so I just plugged it into the original profit equation.
.
Which is this
yeah
They then asked me this final thing:
Max value is just the value of P at s = 1 and t = ln(8/3) right?
Because Lagrange gives us the max/min value?
Yeah I think that's what they want
if it requires you continue with the constraint then yeah
positive profit isn't possible
max is negative, yes
Wut
one sec I might have calculated wrong
P(1, ln(8/3)) = 6000(1)e^(ln(8/3)) - 2000 - 120000
How'd you get that?
so Why'd you get 1986?
ok
Ok nice nice
I can't thank u enough for ur help, rlly do appreciate it. Ur the only person who explains everything so clearly and concisely and correctly everytime
np
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