Lagrange Multipliers Quiz - Library of Math::
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Find the absolute maximum and minimum of the function f(x,y)= x^2 + y^2 subject to the constraint x^4 + y^4 =256.
so far i've gotten
2x = λ4x
2y = λ4y
x^4 + y^4 = 256
So, would I start off the problem with λ = 1/2 and x =0?
Or am I totally off?
We have:
f(x,y) = x² + y²
subject to the constraint:
x⁴ + y⁴ = 256.
We let g(x,y) = x⁴ + y⁴ - 256. Then ∇f = ∂f/∂x i + ∂f/∂y j = 2x i + 2y j and ∇g = ∂g/∂x i + ∂g/∂y j = 4x³ i + 4y³ j. By the Lagrange multiplier method, we have ∇f = λ ∇g at any maximum or minimum value(s) of f for the given constraint, for some non-zero real value λ. Hence,
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2x i + 2y j = ∇f = λ ∇g = λ (4x³ i + 4y³ j) = 4λx³ i + 4λy³ j;
hence,
2x = 4λx³
2y = 4λy³.
The equations give us a number of possibilities that we must examine:
x = 0 and y = 0,
x = 0 and y ≠ 0,
x ≠ 0 and y = 0,
x ≠ 0 and y ≠ 0.
(Analyzing the possibilities is usually the only creative aspect of solving a Lagrange multiplier problem.)
If x = 0 and y = 0, then by subsituting into the constraint, we have:
0⁴ + 0⁴ = 256, a contradiction.
Next, if x = 0 and y ≠ 0, then substituting into the constraint, we have:
0⁴ + y⁴ = 256 ⇒ y = ± 4.
This gives us the points (0,4) and (0,-4) that may provide a maximum or minimum.
Next, if x ≠ 0 and y = 0, then substituting into the constraint, we have:
x⁴ + 0⁴ = 256 ⇒ x = ± 4.
This gives us the points (4,0) and (-4,0) that may provide a maximum or minimum.
Next, if x ≠ 0 and y ≠ 0, we have:
λ = 1/(2x²) = 1/(2y²) ⇒ x² = y² ⇒ x = ± y,
and then substituting into the constraint, we have:
(± y)⁴ + y⁴ = 256 ⇒ 2y⁴ = 256 ⇒ y = ± ⁴√128 = ± 2⁴√8.
This gives us the points (2⁴√8,2⁴√8), (2⁴√8,-2⁴√8), (-2⁴√8,2⁴√8), and (-2⁴√8,-2⁴√8) that may provide a maximum or minimum.
To summarize, we have the following eight points to examine to check for maximum and minimum values of f:
(0,4), (0,-4), (4,0), (-4,0), (2⁴√8,2⁴√8), (2⁴√8,-2⁴√8), (-2⁴√8,2⁴√8), and (-2⁴√8,-2⁴√8).
We plug them into our function f.
We have:
f(0,4) = 0² + 4² = 16,
f(0,-4) = 0² + (-4)² = 16,
f(4,0) = 4² + 0² = 16,
f(-4,0) = (-4)² + 0² = 16,
f(2⁴√8,2⁴√8) = (2⁴√8)² + (2⁴√8)² = 8√8 ≈ 22.627,
f(-2⁴√8,2⁴√8) = (-2⁴√8)² + (2⁴√8)² = 8√8 ≈ 22.627,
f(2⁴√8,-2⁴√8) = (2⁴√8)² + (-2⁴√8)² = 8√8 ≈ 22.627,
f(-2⁴√8,-2⁴√8) = (-2⁴√8)² + (-2⁴√8)² = 8√8 ≈ 22.627.
Hence, the maximum of f, subject to the given constraint, is 8√8 and occurs at the points (2⁴√8,2⁴√8), (2⁴√8,-2⁴√8), (-2⁴√8,2⁴√8), and (-2⁴√8,-2⁴√8) and the minimum of f, subject to the given constraint, is 16 and occurs at the points (0,4), (0,-4), (4,0), and (-4,0).
I should add that the graph of a function f:ℝ²→ ℝ is usually quite easy to visualize above (and/or below) a given constraint on the xy-plane, and in this instance, we see that f is simply a circular paraboloid and that the constraint looks something like a circle that bulges out as it approaches the lines y = x and y = -x; hence, the maximum value of f subject to the given constraint must occur when y = x and y = -x.
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Lagrange Multipliers; Calculus 3?