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Unformatted text preview: Section 13.5 013805.001: If 2 = :r — 3y + 5, then 22 E 1 and 2,, E —3. Hence there is no point on the graph at which the tangent plane is horizontal. Indeed, the graph of z = :c — 3y+ 5 is itself a plane with nonvertical normal
vector (1, —3, ~1), and that’s another reason why no tangent plane is horizontal. (313805.003: If 2 = my + 5, then 2,: = y and 2,, = z. Both vanish at (0, 0),
on the graph of z = I so there is exactly one point
3/ + 5 at WhiCh the tangent Plane is horizontal—{0, 0, ) (313305.020: Given: z=f($,y)::————_l______ 10—2x—4y+x2+y4' Then
222 43,3 4
z ':, = ——__'—“—“_ 2 ———_—“—~—_
f (I y) (10_23+x2 _4y+y4)2 and fy($u y) (lD—2:c+z2—4y+y4)2' Both partials vanish at (I, 1), so there is e xactly one horizontal plane tangent to the graph of z = f (
The point of tangency is (1, 1, :c, y)
This is the highest point on the graph of f. , 013805.035: We will ﬁnd the m
120—the reason in a moment. If :1: z + y + z = 120. So we solve for z, aximum possible product of three nonnegative real numbers with sum
, y, and z are the three numbers, then we are to maximize zyz given
substitute, and maximize f(a:,y)=a:y(120—z—y), 0§x, osy, z+y§l20. Thus by allowing one or two of the numbers to be zero, of the plane—the triangle with two sides on the nonnegative coordinate axes and the third side part of the
graph of y = 120 — z. Write f(:r;, y) = 120mg; — 3:23; — 3:312 and set both partial derivatives equal to zero to
obtain l20y — 22y — yr“) = 0, 12017 ~2xy —;r2 = 0. Because neither 1 nor y is zero~that would mini mize the product, not maximize it—we may cancel to
obtain 12ll—21—y=0, 120—2y—1‘: , and it follows that 22: + y = z + 23/, so that y = x, and then the equation 32
y = 40. it follows that z = 40 as well. The maximum of f( domain, for f(z, y) = 0 there. Hence this lone interior critical
is 40  40 ~ 40 = 64000. 120 yields 1‘ = 40 and
x. y) does not occur on the boundary of its point must yield the global maximum, which ll C13805.036: Let one set of four parallel edges of the box have length x each, another set length y each,
and the third length z each. Then we are to maximize box volume V = zyz given 42: + 4y + 42 = 6; that
is, 2m + 2y + 22 = 3. Solve for z and substitute in V to obtain the function to be maximiZed: A 3—2’ —2' . 3
umbw, osz, osy, z+y§§~ The domain of V is a closed and bounded subset of the plane and V is continuous‘there, so a global
maximum exists. It does not occur on the boundary of the domain because V(:c, y) = 0 there. So it must occur at an interior critical point. When both partials of V are set equal to zero, the resulting equations are 1 2:
:—(3—2:c2y)—my=0, ~2(3—21—Qy)—:ry=0.
It follows that y = z and then, from either of the preceding equations, that y = x = Next, 2 = % as
well, so—because this is the only interior critical point—the maximum volume of such a box is V (é, = g (cubic meters) (313505.044: Suppose that the dimensions of the base and top of the aquarium are a: by 1/ (units are in
inches, cents, etc), that the front and back have dimensions :1: by z, and that the sides have dimensions y by
2. Then the cost of the aquarium will be C = 30mg + 10:52 + 10yz. Solve xyz = 24000 for z and substitute
in the expression for the cost to get the quantity to be minimized: ‘ 240000 + 240000
55 y
The domain of C is not a closed and bounded subset of the plane, but by an argument similar to the one used in Example 7, C(x, y) has a global minimum value that occurs at a critical point in its domain. When
we set both partials of C' equal to zero, we obtain the simultaneous equations C($,y_): +3033], O<:c, 0<y. 30y _ 240200 : 0, 303: _ 240200 2 0
J“ y
and the only real solution of these equations is :2: : 20, y : 20. Thus we have found the location of the
global minimum value of C'(:::, y). The corresponding value of z is 60, so the least expensive aquarium will
have base 20 inches by 20 inches and height 00 inches ( an ideal shape for White Cloud Mountain tropical
ﬁsh). its total cost will be C(20, '20) cents—a rather substantial $360.00. 7 C13805.046: Suppose that (ac, T , z) is the Vertex of the box that lies on the given plane with equation I Z
a b c (a, b, and c are positive constants). We are to maximize the volume V = xyz of the box. Solve the equation
of the plane for z and substitute in the expression for V to obtain V(1:,y)=c:cy(—Ey). OSI, 03y, E+§§L Because the domain of V is a closed and bounded subset of the ryplane and V is continuous there, it has a global maximum. Moreover, the maximum must occur at an interior critical point because V(z, y) is
identically zero on the boundary of its domain. We ﬁrst ﬁnd r, .  _3_E __m_y.§ __ _
“(INN—090 u b) a —ub(b ay 21m) and
:z: y czm crr. Vyriiyl=a< ‘Z‘E)‘—bl=giabbx—2ayl The maximum does not occur when m = 0 or when y = 0, so we need only solve simultaneously ub—uy—21)9:=0 and ab—bx—2ay=0. These equations imply that (1.3] + 21):". = bx + 2m, so that In: = ay. Then substitution of bx for ay in the
equation (11) — bx — 2uy = 0 yields ab—bw—2b1=0; thatis, 12%. By symmetry, the corresponding values of y and z are b
31:12: and 2: win Therefore the maximum value of the box in question is abc abc 333 27' Section 13.9 C13SOQ.001: Given f(:c, y) = 227+y and the constraint g(1:, y) = :1:2 +312 1 = 0, the equation Vf = Avg
yields 2 = 2):: and 1: 2Ay, so that A # 0. Hence 2 a? = 23/. and thus 4y? + y2 = 1 Thus we have two solutions: '2 1 2 1 —~ a, y) = and a, y) = (_—,.¢5, —5v’5 ).
d a 0 Clearly the ﬁrst maximizes f(zr, y) and the second minimizes f(:r, y). So the global maximum value of f(1:, y) is J5— and its global minimum value is —\/5—i Note: The function f is continuous on the circle 2:2 + y2 z 1 and a continuous function deﬁned on a closed
and bounded subset of euclidean space (such as that circle) must have both a global maximum value and
a global minimum value. We will use this argument to identify extrema when possible and without stating
the argument explicitly in various solutions in this section. (313809.011: Clearly f(:r, y, z) = my + 2s is continuous on the closed and bounded spherical surface
g(:r, y, z) = $2 ,‘+— y2 + 32  36 = 0, and hence f has both a global maximum and a global minimum there.
The Lagrange multiplier equation is (y, 1:, 2) = /\(2;r:y 2y, 2:), and hence y = 2Ax, a: = 2/\y, and 2 = 2):.
Now A: = 1, so /\ % (i, If neither 1: nor y is zero, then 1 2:1: 23/
—:——:——:2. A y 1: So 3/2 = 3:2 in this case. If y = x, then z = ‘2, so 2:1:2 + 4 = 36 and we obtain two critical points and the
values f(4, 4, 2) = 20 and f(——4, 4. 2) = ‘20. H? = —x, then 2 =: ~2, and we obtain two critical points and
the values f(4, «4, —‘2) = ~20 and f(—4, 4, —‘2) = ——'20. Finally, if either ofx and y is zero, then so is the
other; :2 = P6, and We obtain two more critical points and the values f(0, 0, 6) = 12 and f(0, 0, ~6) = —12.
Hence the global maximum value of f(;z:, y, z) is 20 and its global minimum value is ~20. (313809.019: We should ﬁnd only one possible extremum because there is no point on the line farthest;
from the origin. We minimize f(;r, y) = m2 + 3/2 given the constraint g(:r, y) = 31' + 4y — 100 = 0. The
Lagrange multiplier method yields (‘21:, 2y) 2 /\(3, 4 ), so that ‘22: = 3A and 2y = 4A. Thus 83'. = 12A = 63;, so that y = Substitution in the constraint yields 3 = 12 and y = 16. Answer: The point on the line y(;r, 3/) =0 closest to the origin is (12, 16). (313809.036: The semiperimeter s = ﬂu: + y + z) of the triangle with sides 9:, y, and z is ﬁxed. We
maximize the square of its area A, A2 = f($, y, z) = s(s — 2:)(5 ~ y)(s — z) (Heron’s formula), sub jeét to the constraint m+y+z = 23; note also that 2:, y, and z are nonnegative. The Lagrange multipler
equations are s(s — y)(s — z) = A, —s(s —— —~ z) = A, and ~ S(s — 1:)(8 — y) = A, and therefore is —w)<s —y) = (s wit» —2) = (s ~y><s—z>. Clearly I < s, y < s, and z < s if the area is to be maximal. Thus
s—zr:=s—y=s—z, 2
andhence x=y=z=§sz The triangle of maximal area is equilateral. It has area 14: 3(8  Hills  y)(s  (313809.038: We maximize and minimize the square of the distance of a point (z, y) of the ellipse from
the origin; thus we maximize and minimize ﬁx: y) = $2 + y2
subject to the constraint g(a;, y) = 1'2 + my + 3/2 — 3 = 0. The Lagrange multiplier equations are
2$=A(2:c+y) and 2y: A(2y+z). (1) Solution 1 (in which we ignore the Suggestion): Clearly A 7é 0. So if neither :5 nor 3/ is zero, we have l_2:n+y_2y+z A‘ 2:. “ 2y ’
y a: 1 —=1 —;
+22 +2y 2312:2552; y2=$2 Case 1: y = x. Then the constraint equation yields 3:52 = 31 so we obtain the two critical points (1, 1),
and(—1,—l). Case 2: y = 1:. Then the constraint equation yields 32 = 3, so we obtain the two critical points (ﬂ, —\/§) and (—x/I’T, Case 3: :r = 0. Then the constraint equation yields y = iﬁ, so we obtain the two critical points (0, and (0, ~\/3_), Case 4: y = 0. In a manner similar to that in Case 3, We obtain the two critical points , O) and (—ﬂ, 0). The values of ﬂat, 3/) at these eight critical points are 2, 2, 6, 6, 3, 3, 3, and 3. Thus the two points on
the ellipse closest to the origin are (1, 1) and (—1, l), each at distance x/i— . The MO farthest from the origin are vﬁ) and (—\/§, ﬂ), each at distance Solution 2 (in which we follow the Suggestion): We write the equations in (l) in the form (2 — 2A)r — Ay = 0,
Art — (2 —2/\)y = 0.
Their solution (17, y) cannot be (0, O), and there must be a nontrivial solution. Hence the determinant of
the preceding system is zero:
—(2 — 2A)2 + A2 = 0;
—4+8A4A2+/\2 :0;
3A2 — 8A + 4 = o;
(A — 2am — 2) = 0.
Case 1: A = Then (1) yields 3 3 so that y = This brings us to Case 1 of the previous solution. Case 2: A = 2. Then (1) yields "
—'21: — 2y 2 0, so that y = —'r.. This brings us to Case 2 of the previous solution. The conclusions in the second solution are exactly the same as in the first solution, but the annoying Cases
3 and 4 of the ﬁrst solution are avoided. ...
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This note was uploaded on 03/01/2008 for the course MATH 51 taught by Professor Staff during the Spring '07 term at Stanford.
 Spring '07
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