Hi! i'm really stuck on this question to do with optimisation of constrained functions. I know i'm meant to use lagrange multipliers method, but i don't know how to start this one of... i'd really appreciate it if someone could please help me out:
Find the maximum and minimum values of:
f(x,y) = 4x + y + y^2
where (x,y) lies on the circle x^2 + y^2 + 2x + y = 1
Find the maximum and minimum values of:
f(x,y) = 4x + y + y^2
where (x,y) lies on the circle x^2 + y^2 + 2x + y = 1
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objective function : f(x,y) = 4x+y+y²
constraint : g(x,y) = x²+ y²+2x+y−1 = 0
We want to find max/min values of f such that g=0.
The method of Lagrange Multipliers says that the solutions are to be found amongst the solutions to the equations ∇f=λ∇g and g=0. λ is a scalar called a Lagrange Multiplier.
In this case these equations are …
4 = λ(2x+2) … (i)
1+2y = λ(2y+1) … (ii)
x²+y²+2x+y−1 = 0 … (iii)
From (ii) (λ−1)(2y+1) = 0. Hence λ=1 or y=−½
If λ=1 then from (i) x=1. Using this in (iii) gives y²+y+2 = 0.
This quadratic has no real solutions so we must have λ≠1.
If y=−½ then from (iii) x²+¼+2x−½−1 = 0 → 4x²+8x−5 = 0
This factors to (2x−1)(2x+5) = 0 so has solutions x=½ or x=−5/2.
λ can be calculated from (i) as 4/3 or −4/3.
So solutions to (i),(ii) & (iii) are …
x=½, y=−½, λ=4/3, f=7/4 and x=−5/2, y=−½, λ=−4/3, f=−41/4
There is a second derivative test to identify the max/min nature of each point but if possible try to use common sense.
The constraint is a circle and the value of f, a quadratic, will vary continuously around the circle and its value will be bounded above and below. So one point must be the maximum of f and the other the minimum of f.
constraint : g(x,y) = x²+ y²+2x+y−1 = 0
We want to find max/min values of f such that g=0.
The method of Lagrange Multipliers says that the solutions are to be found amongst the solutions to the equations ∇f=λ∇g and g=0. λ is a scalar called a Lagrange Multiplier.
In this case these equations are …
4 = λ(2x+2) … (i)
1+2y = λ(2y+1) … (ii)
x²+y²+2x+y−1 = 0 … (iii)
From (ii) (λ−1)(2y+1) = 0. Hence λ=1 or y=−½
If λ=1 then from (i) x=1. Using this in (iii) gives y²+y+2 = 0.
This quadratic has no real solutions so we must have λ≠1.
If y=−½ then from (iii) x²+¼+2x−½−1 = 0 → 4x²+8x−5 = 0
This factors to (2x−1)(2x+5) = 0 so has solutions x=½ or x=−5/2.
λ can be calculated from (i) as 4/3 or −4/3.
So solutions to (i),(ii) & (iii) are …
x=½, y=−½, λ=4/3, f=7/4 and x=−5/2, y=−½, λ=−4/3, f=−41/4
There is a second derivative test to identify the max/min nature of each point but if possible try to use common sense.
The constraint is a circle and the value of f, a quadratic, will vary continuously around the circle and its value will be bounded above and below. So one point must be the maximum of f and the other the minimum of f.
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Try to solve this qn by differentiating the function w.r.t y.