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LAGRANGE mULTIPLIERS

By Rohit Venkat. LAGRANGE mULTIPLIERS. Lagrange Multipliers: A General Definition. Mathematical tool for constrained optimization of differentiable functions Provides a strategy for finding the maximum/minimum of a function subject to constraints. Key Terms.

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LAGRANGE mULTIPLIERS

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  1. By RohitVenkat LAGRANGE mULTIPLIERS

  2. Lagrange Multipliers: A General Definition • Mathematical tool for constrained optimization of differentiable functions • Provides a strategy for finding the maximum/minimum of a function subject to constraints

  3. Key Terms • Gradient – a normal vector to a curve (in two dimensions) or a surface (in higher dimensions) • Lagrange Multiplier – a constant that is required in the Lagrange function because although both gradient vectors are parallel, the directions and magnitudes of the gradient vectors are generally not equal

  4. How to Use the Method of Lagrange Multipliers • Step 1: Form the Lagrangian • Λ(x, y, λ) = f(x, y) + λ(g(x, y) − c) – Lagrangian • f(x, y) – Optimization function • g(x,y) = c – Constraint function • λ– Lagrange multiplier

  5. How to Use the Method of Lagrange Multipliers (Continued) • Step 2: Take the gradient of the Lagrangian and set the equations equal to zero • δΛ/δx = f(x, y)/δx + λ(g(x, y) − c)/δx = 0 • δΛ/δy = f(x, y)/δy + λ(g(x, y) − c)/δy = 0 • δΛ/δλ = g(x, y) − c = 0 • Step 3: Find all critical values by solving the system • Step 4: Evaluate f(x, y) at each set of values to determine maxima/minima

  6. Find the dimensions of the box with largest volume if the total surface area is 96 cm2. • Step 1: • Volume need to be optimized while surface area is constrained • Optimization function: f(x, y, z) = xyz • Constraint function: g(x, y, z) = 2xy + 2yz + 2xz = 96 • Λ(x, y, z, λ) = xyz – λ(xy + yz + xz – 48)

  7. Find the dimensions of the box with largest volume if the total surface area is 96 cm2. • Step 2: • Take the gradient of the Lagrangian and set the equations equal to zero • (1) δΛ/δx = yz– λ(y + z) = 0 • (2) δΛ/δy = xz– λ(x + z) = 0 • (3) δΛ/δz = xy– λ(x + y) = 0 • (4) δΛ/δλ= xy +yz+ xz - 48 = 0

  8. Find the dimensions of the box with largest volume if the total surface area is 96 cm2. • Step 3: • Multiply equation (1) by x, equation (2) by y, and equation (3) by z • (1) xyz – λ(xy + xz) = 0 • (2) xyz – λ(xy + yz) = 0 • (3) xyz – λ(xz + yz) = 0 • (4) xy +yz + xz - 48 = 0 • Set equations (1) and (2) equal to each other • λ(xy + xz) = λ(xy + yz) • λ(xz– yz) = 0 • Either λ = 0 or x = y

  9. Find the dimensions of the box with largest volume if the total surface area is 96 cm2. • Step 3 (Continued): • Set equations (2) and (3) equal to each other • λ(xy + yz) = λ(xz + yz) • λ(xy – xz) = 0 • Either λ = 0 or y = z; so x = y= z • Step 4: • Substitute x for y and z in the constraint function • x2 + x2 + x2 = 48 • 3x2 = 48 • x2 = 16 • x = 4; x = y = z = 4

  10. Find the closest point to the origin on the intersection of the two planes 2x + 3y – 4z = 1 and x + y + 2z = 0. • Step 1: • Form the Lagrangian • Minimize: f(x, y, z) = (x2 + y2 + z2)½ • Subject to: g(z, y, z) = 2x + 3y – 4z – 1 = 0 h(x, y, z) = x + y + 2z = 0 • Lagrangian: Λ(x, y, z, λ, μ) = (x2 + y2 + z2)½ + λ(2x + 3y – 4z – 1) + μ(x + y + 2z)

  11. Find the closest point to the origin on the intersection of the two planes 2x + 3y – 4z = 1 and x + y + 2z = 0. • Step 2: • Take the gradient of the Lagrangian and set the equations equal to zero • (1) δΛ/δx = x/(x2+ y2 + z2)½ + 2λ + μ = 0 • (2) δΛ/δy = y/(x2+ y2 + z2)½ + 3λ + μ = 0 • (3) δΛ/δz= z/(x2+ y2 + z2)½ – 4λ+ 2μ = 0 • (4) δΛ/δλ= 2x + 3y – 4z – 1 = 0 • (5) δΛ/δμ= x + y + 2z= 0

  12. Find the closest point to the origin on the intersection of the two planes 2x + 3y – 4z = 1 and x + y + 2z = 0. • Step 3: • Multiply equations (1), (2), and (3) by (x2 + y2 + z2)½ • (1) x + (x2 + y2 + z2)½(2λ + μ) = 0 • (2) y + (x2 + y2 + z2)½(3λ + μ) = 0 • (3) z + (x2 + y2 + z2)½(–4λ+ 2μ) = 0 • Substitute equations (1), (2), and (3) into equation (5) • –(x2 + y2 + z2)½(2λ + μ) –(x2 + y2 + z2)½(3λ + μ) – 2(x2+ y2 + z2)½(–4λ+ 2μ) = 0 • –2λ–μ– 3λ–μ+8λ– 4μ= 0 • λ = 2μ

  13. Find the closest point to the origin on the intersection of the two planes 2x + 3y – 4z = 1 and x + y + 2z = 0. • Step 3 (Continued): • Substitute λ= 2μ into equations (1), (2), and (3) • (1) x + (x2 + y2 + z2)½(5μ) = 0 • (2) y + (x2 + y2 + z2)½(7μ) = 0 • (3) z + (x2 + y2 + z2)½(–6μ) = 0 • Set equations (1), (2), and (3) equal to each other • x/5 = y/7 = –z/6 • y = 7/5x; z = –6/5x • Substitute y = 7/5x and z = –6/5x into equation (4) • 2x + 3(7/5x) – 4(–6/5x) – 1 = 0 • x = 1/11

  14. Find the closest point to the origin on the intersection of the two planes 2x + 3y – 4z = 1 and x + y + 2z = 0. • Step 3 (Continued): • Plug x = 1/11 back in for y and z • y = (7/5)x = (7/5)(1/11) = 7/55 • z = (–6/5)x = (–6/5)(1/11) = –6/55 • Closest Point: (1/11, 7/55, –6/55)

  15. Works Cited • http://en.wikipedia.org/wiki/Lagrange_multipliers • http://mathworld.wolfram.com/LagrangeMultiplier.html • http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html#Method • http://math.mit.edu/classes/18.02/notes/lecture13-09.pdf

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