DOC PREVIEW
MIT 18 02 - LECTURE NOTES

This preview shows page 1 out of 2 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

18.02A Topic 28: Max-min problems, least squares.Author: Jeremy OrloffRead: TB: 19.7, SN: LSStandard calculus question:Given z = f(x, y) where are the (relative) maxima and minima?Answer: At the points where ∇f = h0, 0i.I.e.∂f∂x= 0 and∂f∂y= 0.Critical points: Such points are called critical points.Example: z = x2+ y2∇z = h2x, 2yi ⇒ only critical point is (0, 0). Clearly a minimum.Example: z = −x2− y2∇z = h−2x, −2yi ⇒ only critical point is (0, 0).Clearly a maximum.Example: z = y2− x2∇z = h−2x, 2yi ⇒ only critical point is (0, 0).Clearly neither maximum or minimum.Critical point = horizontal tangent planeEquation of tangent plane at (x0, y0, z0) (z0= f(x0, y0)) is(z − z0) =∂f∂x(x0,y0)(x − x0) +∂f∂y(x0,y0)(y − y0).At a critical point this becomes: z − z0= 0 = horizontal plane.Example: Making a box.Sides double thick, bottom triple thick, no top, volume = 3.What dimensions use the least amount of cardboard?Area of one side = yz, two sides, double thick ⇒ cardboard used = 4yz.Area front (and back) = xz, single thick ⇒ cardboard used = 2xz.Area bottom = xy, triple thick ⇒ cardboard used = 3xy.Total cardboard used = w = 4yz + 2xz + 3xy.Volume = 3 = xyz ⇒ z =3xy⇒ w =12x+6y+ 3xyCritical points: wx= −12x2+ 3y = 0, wy= −6y2+ 3x = 0wx= 0 ⇒ y =4x2wy= 0 ⇒ −616x4+ 3x = 0⇒ x = 2, y = 1, z = 3/2.Clearly a minimum. (physically we know it must have a minimum somewhere, itcan’t be on the boundary (axes) because w is infinite there ⇒ only critical pointmust be a minimum.(continued)118.02A topic 28 2Least squares You should read §LS in the notes.Start with data points: (x1, y1), (x2, y2), . . . , (xn, yn).Try to fit a line, y = ax + b, to the data.The ’squared error sum’ is E(a, b) =Xi(yi− (axi+ b))2.The least squares fit is the line that minimizes E.To find the minimum we set∂E∂a= 0 and∂E∂b= 0.∂E∂a=X−2xi(yi− (axi+ b)) = 2Xiax2i+ bxi− xiyi= 0.∂E∂b=X−2xi(yi− (axi+ b) = 2Xiaxi+ b − yi= 0.This gives the least squares equations for a line:⇒ aXix2i+ bXixi=XixiyiaXixi+ b n =XyiExample: Use least squares to fit a line to the following data: (0,1), (2,1), (3,4).answer: In our case, (x1, y1) = (1, 1), (x2, y2) = (2, 1) and (x3, y3) = (3, 4).⇒Px2i= 13,Pxi= 5, n = 3,Pxiyi= 14,Pyi= 6.⇒ 13a + 5b = 14; 5a + 3b = 6 ⇒ a = 6/7 and b = 4/7.The least squares line has equation y =67x +47.Example: For the same points as above, use least squares to fit a parabola.answer: A parabola has the formula y = ax2+ bx + c.Squared error = E(a, b, c) =X(yi− (ax2i+ bxi+ c))2.The least squares fit minimizes E ⇒∂E∂a=∂E∂b=∂E∂c= 0:⇒ aXx4i+ bXx3i+ cXx2i=Xx2iyiaXx3i+ bXx2i+ cXxi=XxiyiaXx2i+ bXxi+ c n =XyiPlugging in our points:Px4i= 97,Px3i= 35,Px2i= 13,Pxi= 5, n = 3.Px2iyi= 40,Pxiyi= 14,Pyi= 6.⇒ 97a + 35b + 13c = 4035a + 13b + 5c = 1413a + 5b + 3c = 6.Solving ⇒ a = 1, b = −2, c = 1.The least squares parabola has equation y = x2− 2x + 1.Note: for 3 points the fit is perfect.xy• ••Least squares lineand


View Full Document

MIT 18 02 - LECTURE NOTES

Documents in this Course
Vectors

Vectors

1 pages

Exam 1

Exam 1

2 pages

Load more
Download LECTURE NOTES
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view LECTURE NOTES and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view LECTURE NOTES 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?