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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 9
lVlax/ min problems. Least squares approximation February 24 Last time: Partial derivatives. Tangent planes. Tangent plane approximation
Today: Max/min problems. Least squares approximation Reading Material: From Simmos 19.7. From the Lecture Notes LS. In this lecture we will learn how to ﬁnd the local maximum and minimum values of multivariable
functions. To get an idea of a typical application let’s consider the following example: Exercise 1. [Post Ofﬁce Problem] Suppose you are moving to Europe for a year and you are
allowed to use the post oﬂ‘ice to sencl only ONE boat The post oﬁice has some restrictions on the
size of homes that one can semi: Size limit: length+ girth g 9 feet. Here ’ b L = Length
‘W = Width
D = Depth L and W 2(W' + D) = girth. Of course you want to use the largest possible how, so you want to mowimize the uolume function by
using the largest possible value given. by the size limit: V : LWD
L.+ 2(W + D) = 9
From the second equation. we unite L in terms of W and D.
L:9—2(W+D)
and we replace it in V:
V = LWD = [9  2(W + D)]WD : 9WD — Ell/ED — Elli/D2.
momimize So the ﬁnal question is ﬁnding the mamimum value of V(ltV,D) and the corresponding maximum
point (WlnDu) (and as a consequence L”) where this maximum is achieved. 2 Max, Min and Critical Points 2D: Consider the one variable function y = ﬁr), then we know that the (local) max/min values for f occur only at the points where the tangent line
is horizontal, that is where $1 = 0. 3D: Consider now the two variable function 5 = '(:1:, y) (imagine a landscape), 2: §("',. Dc) ! i
4 05159) ({14'3' ) then the max/min occur only at those points where the tangent planes are horizontal, that is the
planes are parallel to the Icyplane. This means that the equation of these planes can be written as
2 = c that is at these points g—: = O and 37'; = 0; Deﬁnition 1. Given a function z = f(:x,y) all points (Imyu) such that (.34) ' =0 and (g) :0
'7‘ (meme) 6y (1110.110) are called Critical Points. Note: If ($0,110) is a point of either local Max or Min value for a function then (3:0,yu) must he a
critical point. But the reverse is not true in the sense that a point ($0,190) may be a critical point,
but the function may not have either local Max or Min there. See teh example in page 3 and 4. Now we know that to solve the Post Office Problem we need to look for critical points for the function
VUJV, D): set §L_a_0
aw ‘ an ‘ ' [Q Solve for l/V, D. 8V .,
—...._ = _ 7 k 2 "' =
(9er 9D 414 D D 0
13(2) —— 4W — 2D) = 0
9 — 41/ " — 21) = 0
since D = 0 certainly would not be a possible solution for us! On the other hand / (
E :914/ ; :uVD — 2W3 = 0 6D
III/(9 — 4W — 2D) : 0
9 ~ 40 — 2W 2 0
again since 31V : 0 certainly would not be a possible solution for us! \lVe get the system 9—4W—2D=O
9—4D—21IV=D and from here subtracting from the ﬁrst the second equation —2W + 2D : 0 +=p W = D
9 — =10 — 2W = 0 and finally
W = D = 3/2. This is the only critical point where a maximum is possible (the minimum would he at. (O, O) that
we discarded). Going back to the problem we also obtain L=9—2(W+D):3 and your box will be 3 X 3/2 x 3/2 x 7 cubic feet!
Quetion: Are critical points ALWAYS max or Min points? Answer: N 0! 213: Take for example 3; = 3:3 at U 7 Clearly hence 0 is a critical point, but it is only an inﬂection point, the tangent line doesn't stay just in one
side of the graph for points “near” 0. 3D: Now consider the function 3 = 3:2 — if" at the point (0, 0) Clearly (03> +0)
6"” (0,1)) . By ((1.0) hence (0,0) is a critical point, but it is only a saddle point, the tangent plane doesn’t stay just in
one side of the graph for points “near” (0,0). 3 Least squares approximation Suppose that in your lab or in your ﬁeld work you gathered a collection of data that can be repre—
sented on a two dimensional plane: 7
3:».er 4 3069‘ we” D : {(351,101), ‘ (513”.ynll
{xl,3:) . and you notice they ”cluster" near a line. Can you ﬁnd a line y = me: + b that best represent this
set D of data? YA JakakPb 7 E; " .6": (names—7;] For each data point (2:1, y:) the error we make in replacing yi with the value of the line at the point
rt'i, that is (my:1 + b), is ei : lyi — ('mrt'i + b).
Then the total error is n.
E1 4 82 + +8“ =2 81'.
i=1 The goal is to ﬁnal the unknown m and b by minimizing this error. It turned out that ['or computa
tional reasons it is better to actually consider the total error H
E:e}?+—e':_i§ve,2l =25?
i=1
and this doesn’t really change the problem. So our goal is to minimize E as function of m and I}.
Since 63 = {yi w (maxi + 45)}2 it follows that
1!
E = 2m — (Trtﬂii + 5))1,
1=1 where (:t'i‘yi) are given for 'i 2 1, r..,'n. 11nd (m,b) are the variables. We look for the critical points
of E (LE _ 0 _ 8_E
0m _ _ 8b.
UE ”
— t 2':—"'1 —5'::
0m E (y (me + b))( L ) U
kayyi + Z owl2 + Zim : 0
and ﬁnally
m21‘?+bZﬂli=er1yi. (3.1}
On the other hand
SE
at = 22% — (mt1 +b)){—1) : 0 21—1}: + mmi +11) = 0 ~Zyi + 27113:; + 21): U
m Z 531+ nb = Egg. (3?) Now if we deﬁne the following overage values and from here Tl _ l
.1: : avg. of mi 2 ~ 1';
n r
L=l
H
_ 1
3}:an of:z = —Zyi
n ‘
1:1
H
“3 2 1 a
‘ : (mg. of h 7 ~— 1:;
1'1
1:]
7i
__ . 1
rt = avg of mzyl : —— 1.19:
n
i=1 and we divide both (3.1) and (3.2) by n, we obtain the following system for the critical points: ml‘2 + bi‘ my mi+b : (Cir Notice that this system is equivalent to the matrix system (f i)( EL)=( ?) <=>AX=B Exercise 2. Given points (1,2) (2,6) (3,4) ﬁnd best ﬁt line: _ _ 2 + 12 + 12 26
t = 4 .y : z —
3 3
Then the system becomes 14 26 — + 2b — — 3 m 3 2m + b = 4 and the unique solution is which, gives (he line y = .’L' + 2 (LS best ﬁt: 3 Study Guide 1. Answer the following questions: a Can you summarize the geometric "reason (tnwot'uing tangent lines or tangent planes) why tocat
mom and min are obtained at critical points? a How would yon find the best ﬁt parabola of equation y = (1:32 + b? a Can you ﬁnd another example in SD for which a critical point is neither a may: or min point?...
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