HELP ASAP Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given?
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Answer:
We want to optimize f(x, y) = e^(xy), subject to g(x, y) = x^5 + y^5 = 64. By Lagrange Multipliers, ∇f = λ∇g ==> <ye^(xy), xe^(xy)> = λ<5x^4, 5y^4>. Equating like entries, ye^(xy) = 5λx^4 and xe^(xy) = 5λy^4. Hence, xye^(xy) = 5λx^5 = 5λy^5. We can't have λ = 0, since this implies that (x, y) = (0, 0), which does not satisfy g. So, x^5 = y^5 ==> x = y. Substituting this into g yields (x, y) = (2, 2). ==> f(2, 2) = e^4 is the maximum. (Why? Note that (0, 64^(1/5)) also satisfies g, but f(0, 64^(1/5)) = 1 < e^4.) I hope this helps!
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Other answers
We want to optimize f(x, y) = e^(xy), subject to g(x, y) = x^5 + y^5 = 64. By Lagrange Multipliers, ∇f = λ∇g ==> <ye^(xy), xe^(xy)> = λ<5x^4, 5y^4>. Equating like entries, ye^(xy) = 5λx^4 and xe^(xy) = 5λy^4. Hence, xye^(xy) = 5λx^5 = 5λy^5. We can't have λ = 0, since this implies that (x, y) = (0, 0), which does not satisfy g. So, x^5 = y^5 ==> x = y. Substituting this into g yields (x, y) = (2, 2). ==> f(2, 2) = e^4 is the maximum. (Why? Note that (0, 64^(1/5)) also satisfies g, but f(0, 64^(1/5)) = 1 < e^4.) I hope this helps!
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