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# MATH 2010 - Multivariable Calculus and Matrix Algebra (Herron) - Week 3 Notes (S16) MATH 2010

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This 4 page Class Notes was uploaded by creask on Friday February 12, 2016. The Class Notes belongs to MATH 2010 at Rensselaer Polytechnic Institute taught by Isom Herron in Spring 2016. Since its upload, it has received 39 views. For similar materials see Multivariable calculus and matrix algebra in Mathematics (M) at Rensselaer Polytechnic Institute.

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Date Created: 02/12/16

MATH 2010 - Multivariable Calculus & Matrix Algebra Professor Herron - Rensselaer Polytechnic Institute Week 3 (2/8/16 - 2/12/16) Important: These notes are in no way intended to replace attendance in lecture. For best results in this course, it is imperative that you attend lecture and take your own detailed notes. Please keep in mind that these notes are written speci▯cally with Professor Herron’s sections in mind, and no one else’s. The 2nd Derivatives Test To remember the formula for D, write it as a determinant: ▯ ▯ ▯fxx fxy▯ 2 D = ▯f f ▯ = f xx yy ▯ (fxy yx yy • This is called a Hessian determinant. Example 1 Consider the three points: (0;2);(1;▯5);(2;4). Compute and solve for y: m(x + x + x ) + 3b = y + y + y 1 2 3 1 2 3 and 2 2 2 m(x +1x + 2 ) +3b(x + x 1 x )2= x y3+ x y1 1x y :2 2 3 3 • Substitute in the points: x y1 1x y +2 2y = 3 3 5 + 8 = 3 • m(0 + 1 + 2) + 3b = 2 ▯ 5 + 4 = 1 ! 3m + 3b = 1 • m(0 + 1 + 4) + b(0 + 1 + 2) = 3 ! 5m + 3b = 3 • Solve for m by subtracting: 2m = 2 ! m = 1 ▯2 { This in turn means that b = 3 2 { Thus, y = x ▯ 3 iClicker • A smooth function z = f(x;y) satis▯es the conditions f (a;b) =x0 and f (a;b) = 0y Then f has a horizontal tangent at (0;0). 2 • The point (1;0) is a solution to rf(x;y) = h0;0i for the function f(x;y) = x ▯ 2x + y ▯ 1. Using the 2nd Derivatives Test, determine what extrema (1;0) is. 1 { f xx = fyy = 2 and f xy = 0 { So (2)(2) ▯ 0 = 4, making (1;0) a local minimum. Global/Absolute Extrema • For maxima, the global maximum is the largest maximum of a function. • Likewise, the global minimum is the largest minimum of a function. Bounded Sets • On a at surface, R , a bounded set is a ▯nite set contained within some disk. 3 • A bounded set in space, or R , can be contained in a sphere. • A disk of arbitrarily small radius contains points in the set and is in the set if the disk is on the boundary of the set. • An open set has every boundary point not in the set. Extreme Values Theorem (2-Variable Functions) 2 If f is continuous on a closed, bounded set D in R , then f attains a global max f(x ;y ) 1 1 and a global min f(x ;y2) 2t some points (x ;y )1x 1y ) 2n 2. Example 2 2 2 2 2 Find the global max and global min for z = x + 2y within and on x + y = 1. • z = 2x and z = 4y ! z = z at the point (0;0). x y x y • zxx = 2;z yy = 4;z xy = 0 ! (0;0) is a local min ! global min. • Global max lies on the circle. { z = x 2+ 2(1 ▯ x ) = 2 ▯ x , for ▯1 ▯ x ▯ 1. { The global max occurs where x = 0 (z = 2) max Lagrange Multipliers (2 Variables) • rf = ▯rg (should be pointing in the same direction) • rf is perpendicular to the level curve. { f(r(t)) is equal to a constant. @ { @tf(r(t)) = rf ▯ r = 0. { g is constant. 2 Example 3 Find the max or min of z = x + 2y on a circle x + y = 1 using Lagrange multipliers. • Use rf = ▯rg. • rf = h2x;4yi and rg = h2x;2yi • Solve: 2x = ▯2x, 4y = ▯2y; x + y = 1 ! constant. • Main idea: rf is parallel to rg, so solve simultaneously. { In the ▯rst equation, x = 0 or ▯ = 1 { In the second equation, y = 0 or ▯ = 2 { By the third equation: ∗ If ▯ = 1, then y = 0 and x = 1 ! x = ▯1 ∗ If ▯ = 2, then x = 0 and y = 1 ! y = ▯1 • Therefore, f has a global max or min on D. Lagrange Multipliers (3 Variables) • Instead of level curves, we now have level surfaces. • The max of the function f(x;y;z) = c is tangent to the level surface g(x;y;z) = k. • rf = ▯rg { f(x(t);y(t);z(t)) = c d 0 { dt = 0 ! rf ▯ r = 0 Example 4 A rectangular box is to be made from 12 m of cardboard. Find the max volume when the box has no lid. • Vmax = xyz, where 2xz + 2yz + xy = 12. 12▯xy xy(12▯xy) • Solving for z: z = 2x+2y, so V = 2x+2y . • One way to ▯nd V would be to use the quotient rule. Instead, use Lagrange multipliers. • rV = ▯rg, where g = 2xz + 2yz + xy • rV = ▯h2z + y;2z + x;2x + 2yi • constraint: 2xz + 2yz + xy = 12 • xyz 6= 0;x > 0;y > 0;z > 0 ! ▯ = 6 0 • xyz = ▯(2xz + xy) = ▯(2yz + xy) = ▯(2xz + 2yz) 3 { 2xz +▯xy = 2yz▯+ xy ! x = y { 2xz +▯2yz ▯ 2yz + xy ! 2z = y • Subbing into the constraint: 4z + 4z + 4z = 12 2 { 12z = 12, so z = 1. { Thus, x = y = 2. 3 • xyz = 4 m It is possible that you will need to solve the max or min of u = f(x;y1z) when g(x;y;z) = k and h(x;y;z) = k ; in other words, 2 constraints. 2 • In this case, use 2 Lagrange multipliers: 8 <rf = ▯rg + ▯rh g(x;y;z) = k > 1 : h(x;y;z) =2k • In this case, ▯ and ▯ are constants (Lagrange multipliers) End of Document 4

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