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16bf10lec4

# 16bf10lec4 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

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Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon Math 16b Question, recalled Last week, we considered the problem of maximizing the area enclosed by a rectangular fence subject to having a budget of \$ 600. Here, the cost of the red side is \$ 7 per foot while the other sides cost \$ 5 per foot. We solved the problem by using the equation 12 y + 10 x = 600 to solve for y as a function of x and then solved the corresponding one variable optimization problem. Thomas Scanlon Math 16b Question: the example turned on its head As before, the cost of the red side is \$7 per foot while the other sides cost \$ 5 per foot. Now, I would like to know the value of x and y which minimize the cost subject to the constraint that the area is 30 square feet. Thomas Scanlon Math 16b Solution using Lagrange multipliers This time we wish to minimize C ( x , y ) := 12 y + 10 x subject to the constraint that xy = 30. In this case, it would be fairly easy to convert the problem to a one variable optimization problem, but for sake of illustration we shall use Lagrange multipliers. We set G ( x , y ,λ ) := 12 y + 10 x + λ ( xy- 30 ) and compute ∂ G ∂ x = 10 + λ y , ∂ G ∂ y = 12 + λ x and ∂ G ∂λ = xy- 30 Thomas Scanlon Math 16b Solution, continued Setting all of the rst partial derivatives of G equal to zero, we have 10 + λ y = 0, 12 + λ x = 0 and xy- 30 = 0. From the rst two equations we conclude- 10 / y = λ =- 12 / x , or by taking reciprocals and additive inverses,...
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16bf10lec4 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

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