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# Calculus help please - Lagrange Multipliers?

Use Lagrange multipliers to find the given extremum. Assume that x and y are positive.

Minimize f(x, y) = x^2 − y^2

Constraint: x − 6y + 105 = 0

Minimum of f(x, y,)= ? at (x, y)= ( ? )

Thank you :)

### 1 Answer

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- kbLv 77 years agoFavorite Answer
We want to minimize f(x, y) = x^2 - y^2 with constraint g(x,y) = x - 6y + 105 = 0.

By Lagrange Multipliers, ∇f = λ∇g.

==> <2x, -2y> = λ<1, -6>

==> λ = 2x = -2y/-6, by equating like entries

==> y = 6x.

Substituting this into g:

x - 6*6x + 105 = 0

==> x = 3.

So, the critical point is (x, y) = (3, 18).

Finally, f(3, 18) = 9 - 324 = -315 is the minimum, since for instance (105, 0) also satisfies g,

but f(105, 0) > -315.

I hope this helps!

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