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Sagot :
The minimum and maximum values are (16/9)√3 and -(16/9)√3.
According to the statement
we have given that the function
First of all, since the constraint's graph is a circle, which is a closed loop, and f is continuous in [tex]R^2[/tex], there must exist a constrained global maximum and minimum value.
Lagrange Multipliers:
f(x,y) = xy^2
g(x,y) = x^2 + y^2
We want ∇f = λ∇g so we get the following system of equations
1. y^2 = 2λx
2. 2xy = 2λy
3. x^2 + y^2 = 4 ← The constraint.
Equation 2 implies that y = 0 or λ = x
We can ignore y = 0, since that will make f(x,y) = 0 and clearly f(x,y) takes on both positive and negative values subject to the constraint.
Plugging in the alternative, λ = x to equation 1 gives y^2 = 2x^2.
Plugging this into the constraint gives 3x^2 = 4 so that x^2 = 4/3 and y^2 = 8/3
Taking square roots gives
x = ±√(4/3) = ±(2/3)√3
y = ±√(8/3) = ±(2/3)√6
At the points < (2/3)√3 , ±(2/3)√6 >, f(x,y) = (16/9)√3 ← Maximum
At the points < -(2/3)√3 , ±(2/3)√6 >, f(x,y) = -(16/9)√3 ← Minimum
So, The minimum and maximum values are (16/9)√3 and -(16/9)√3.
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