Use Lagrange multipliers to determine the maximum value of f(x, y) = 2x − 3y subject to the constraint
x^2 + 2y^2 = 4.
x^2 + 2y^2 = 4.
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Let f(x,y) = 2x - 3y subject to g(x,y) = x^2 + 2y^2 = 4.
By Lagrange Multipliers, ∇f = λ∇g.
==> <2, -3> = λ<2x, 4y>.
Equate like entries:
2 = 2λx and -3 = 4λy
==> x = 1/λ and y = -3/(4λ).
Plug this into g:
(1/λ)^2 + 2(-3/(4λ))^2 = 4
==> 1/λ^2 + 9/(8λ^2) = 4
==> λ^2 = 17/32 = 34/64.
==> λ = ±√34 / 8.
Hence, we have two critical points
(±8/ √34, ∓6/ √34).
Plug these points into f:
f(8/ √34, -6/ √34) = √34 <---Maximum
f(-8/ √34, 6/ √34) = -√34 <---Minimum
I hope this helps!
By Lagrange Multipliers, ∇f = λ∇g.
==> <2, -3> = λ<2x, 4y>.
Equate like entries:
2 = 2λx and -3 = 4λy
==> x = 1/λ and y = -3/(4λ).
Plug this into g:
(1/λ)^2 + 2(-3/(4λ))^2 = 4
==> 1/λ^2 + 9/(8λ^2) = 4
==> λ^2 = 17/32 = 34/64.
==> λ = ±√34 / 8.
Hence, we have two critical points
(±8/ √34, ∓6/ √34).
Plug these points into f:
f(8/ √34, -6/ √34) = √34 <---Maximum
f(-8/ √34, 6/ √34) = -√34 <---Minimum
I hope this helps!
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I recommend the writeup here: http://www.slimy.com/~steuard/teaching/t…