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DEPARTMENT OF MATHEMATICS, IIT - GUWAHATI
Even Semester of the Academic Year 2011- MA 102 Mathematics II Problem Sheet 3: Critical points, maxima and minima, Lagrange’s multipliers and double integrals. Instructors: Dr. J. C. Kalita and Dr. S. Bandopadhyay
- Find the local maximum and minimum values and saddle point(s) of the functions:
(a) f (x, y) = x^2 + y^2 + x^2 y + 4 (b) f (x, y) = 4xy − x^4 − y^4 (c) f (x, y) = sin x cosh y
(d) f (x, y) = x + 2y +
x
− y^2.
Solution: (a) Since f (x, y) = x^2 + y^2 + x^2 y + 4, fx(x, y) = 2x + 2xy and fy (x, y) = 2y + x^2. The critical points are given by the following equations: fx(x, y) = fy (x, y) = 0, which gives x(y + 1) = 0 and 2y = −x^2 , ⇒ (0, 0) and (
2 , −1) are the three critical points. D(x, y) = fxxfyy − f (^) xy^2 = (2 + 2y)(2) − (2x)^2 = 4(1 + y − x^2 ). At (0, 0), D(0, 0) = 4 > 0, and fxx(0, 0) = 2 > 0 hence (0, 0) is a point of local minimum. Since D(
2 , −1) = D(−
2 , −1) are saddle points of f.
(b) Since f (x, y) = 4xy − x^4 − y^4 , fx(x, y) = 4y − 4 x^3 and fy(x, y) = 4x − 4 y^3. The critical points are given by the following equations: fx(x, y) = fy (x, y) = 0, which gives y = x^3 and x = y^3 , or x = x^9 , ⇒ x(x^4 − 1)(x^4 + 1) = 0. (1) The only real roots to (1) are x = 0, +1, −1, ⇒ the critical points are (0, 0), (1, 1), (− 1 , −1). D(x, y) = fxxfyy − f (^) xy^2 = (− 12 x^2 )(− 12 y^2 ) − 42 = 144x^2 y^2 − 16. At (0, 0), D(0, 0) = − 16 < 0, hence (0, 0) is a saddle point. At (1, 1) and (− 1 , −1), D(1, 1) = D(− 1 , −1) > 0, and fxx(1, 1) = fxx(− 1 , −1) = − 12 hence (1, 1) and (− 1 , −1) are both points of local maxima.
(c) fx(x, y) = cos x cosh y and fy(x, y) = sin x sinh y. The critical points are given by the following equations: fx(x, y) = fy (x, y) = 0, which gives x =
(2n + 1)π 2
, y = 0 for n = 0, ± 1 , ± 2 , ....... fxx(x, y) = − sin x cosh y, fyx(x, y) = cos x sinh y, fyy (x, y) = sin x cosh y.
D
( (2n + 1)π 2
) = fxx(0, 0)fyy(0, 0) − (fxy (0, 0))^2 < 0, hence
( (2n + 1)π 2
) for n = 0, ± 1 , ±2 are all saddle points of f.
(d) Since f (x, y) = x + 2y +
x
− y^2 , fx(x, y) = 1 − 4 x−^2 and fy (x, y) = 2 − 2 y, (for x 6 = 0). The critical points are given by the following equations: fx(x, y) = fy (x, y) = 0, which gives x = ±2, y = 1. ⇒ the critical points are (2, 1), (− 2 , 1). D(x, y) = fxxfyy − f (^) xy^2 = ( (^) x^83 )(−2) − 0 , ⇒ D(2, 1) < 0 and D(− 2 , 1) > 0, and fxx(− 2 , 1) < 0. Hence (2, 1) is a saddle point and (− 2 , 1) is a point of local maximum.
- Find the absolute maximum and minimum values of f (x, y) = 4xy^2 − x^2 y^2 − xy^3 on the set D where D is the closed triangular region in the xy-plane with vertices (0, 0), (0, 6) and (6, 0).
Solution: Since f (x, y) = 4xy^2 − x^2 y^2 − xy^3 , fx(x, y) = 4y^2 − 2 xy^2 − y^3 , fy (x, y) = 8 xy − 2 x^2 y − 3 xy^2. The critical points are given by the following equations: fx(x, y) = fy (x, y) = 0, which gives y^2 (4 − 2 x − y) = 0 and xy(8 − 2 x − 3 y) = 0. Hence the critical points are given by the conditions: (i)y = 0, x = 0, (0, 0) on the boundary of D (ii) y = 0, 8 − 2 x = 0, which gives (4, 0) on the boundary of D (iii)4 − 2 x − y = 0, x = 0, which gives the point (0, 4) on the boundary of D (iv) 4 − 2 x − y = 0, 8 − 2 x − 3 y = 0 which gives the point (1, 2), which is in the interior of the triangle D (v) y = 0, 4 − 2 x = 0 which gives (2, 0) which is on the boundary of D.
fxx(x, y) = − 2 y^2 , fxy (x, y) = 8y − 4 xy − 3 y^2 , and fyy (x, y) = 8x − 2 x^2 − 6 xy.
H =
.
Since △ 1 = fxx(0, 0 , 0) = 10 > 0, △ 2 = fxx(0, 0 , 0)fyy(0, 0 , 0) − (fxy(0, 0 , 0))^2 = 64 > 0 and △ 3 = |H| = 600 > 0, hence (0, 0 , 0) is a point of local minimum.
(b) f (w, x, y, z) = wx + 2xy + 3yz − w^2 − 2 x^2 − 3 y^2 − 4 z^2 fw(x, y, z, w) = x − 2 w, fww(x, y, z, w) = − 2 , fxw(x, y, z, w) = 1, fyw(x, y, z, w) = 0 , fzw(x, y, z, w) = 0. fx(x, y, z, w) = w+2y− 4 x, fxx(x, y, z, w) = − 4 , fyx(x, y, z, w) = 2, fzx(x, y, z, w) = 0 fy(x, y, z, w) = 2x + 3z − 6 y, fyy(x, y, z, w) = − 6 , fzy(x, y, z, w) = 3 fz (x, y, z, w) = 3y − 8 z, fzz (x, y, z, w) = − 8. Since fw(x, y, z, w) = fx(x, y, z, w) = fy (x, y, z, w) = fz (x, y, z, w) = 0, at (0, 0 , 0 , 0) hence it is a critical point of f and the Hessian matrix at (0, 0 , 0 , 0) is given by
H =
.
Therefore △ 1 = − 2 < 0, △ 2 = 7 > 0, △ 3 = − 34 < 0, △ 4 = 337 > 0. Since (−1)k△k > 0 for all k = 1, 2 , 3 , 4 , (0, 0 , 0 , 0) is a local maximum point of f.
- Find the points on the surface z^2 = xy + 1 that are closest to the origin.
Solution: The square of the distance of any point of the given surface from the origin is given by: f (x, y) = x^2 + y^2 + z^2 = x^2 + y^2 + xy + 1 To find the absolute minimum of f. The critical points are given by the following equations: fx(x, y) = fy(x, y) = 0, which gives 2x + y = 0 = 2y + x, the only solution to this system is (0, 0). D(0, 0) = fxx(0, 0)fyy(0, 0) − (fxy(0, 0))^2 = 2 × 2 − 1 = 3 > 0 and fxx(0, 0) = 2 > 0, hence (0, 0) is a point of local minimum. Since (0, 0) is the only critical point so (0, 0) is the point of absolute minimum of f.
- Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid 9x^2 + 36y^2 + 4z^2 = 36.
Solution: Since the ellipsoid is centered at (0, 0 , 0), clearly the rectangular box of maximum volume should also be centered at the origin and if (x, y, z) gives a cor- ner of the rectangle in the first octant, which touches the ellipsoid then clearly the other corners are (−x, y, z),(x, −y, z),(x, y, −z),(x, −y, −z),(−x, −y, z),(−x, y, −z) and (−x, −y, −z). Then the volume of the rectangular box is given by V = 8xyz and the point (x, y, z) should satisfy the condition 9x^2 + 36y^2 + 4z^2 = 36. So the problem is to max 8xyz or max cx^2 y^2 z^2 (c > 0) subject to 4z^2 = 36 − 9 x^2 − 36 y^2. or max x^2 y^2 (36 − 9 x^2 − 36 y^2 ). The critical points are given by the following conditions:
fx(x, y) = (2x)y^2 (36 − 9 x^2 − 36 y^2 ) + x^2 y^2 (− 18 x) = 0 or xy^2 (2 − x^2 − 2 y^2 ) = 0 and fy(x, y) = (2y)x^2 (36 − 9 x^2 − 36 y^2 ) + x^2 y^2 (− 72 y) = 0 or yx^2 (4 − x^2 − 8 y^2 ) = 0. The critical points satisfy either of the following conditions: (i) x = 0 (ii) y = 0 (iii) x 6 = 0, y 6 = 0, 2 − x^2 − 2 y^2 = 0 and 4 − x^2 − 8 y^2 = 0, which gives y = ± √^13 and
x = ± √^23.
Since for critical points of the type (i), (ii), V = 0, we need to check for solutions of type (iii) if it maximizes V. (**) Since D(x, y) = fxx(x, y)fyy(x, y) − fxy(x, y)^2 > 0, and fxx(x, y) < 0 at (± √^23 , ± √^13 ) these points give the (x, y) coordinates of the corner points of the in- scribed rectangular solid of maximum volume. Since the z coordinates of the solid are ±
3, and the volume of the required solid is 8 × √^23 × √^13 ×
(**) Aliter: Since the existence of such a solid is guaranteed ( maximizing a con- tinuous function x^2 y^2 z^2 in a closed bounded region) so the critical points other than those given by x = 0, y = 0 ( which gives the minimum value of x^2 y^2 z^2 ) must necessarily maximize x^2 y^2 z^2.
Aliter: Using lagrange’s multipliers to solve the above problem we get: max x^2 y^2 z^2 subject to g(x, y, z) = 4z^2 − 36 + 9x^2 + 36y^2 = 0. Let λ such that ∇f (x, y, z) = λ∇g(x, y, z) ⇒ (2xy^2 z^2 , 2 yx^2 z^2 , 2 zy^2 x^2 ) = λ(18x, 72 y, 8 z) (1)
- Use Lagrange multipliers to find the global minimum and maximum values of the functions subject to the given constraint(s)
(a) f (x, y) = 4x + 6y; x^2 + y^2 = 13
(b) f (x 1 , x 2 ,... , xn) = x 1 + x 2 + · · · + xn; x^21 + x^22 + · · · + x^2 n = 1
(c) f (x, y) = e−xy; x^2 + 4y^2 ≤ 1.
Solution: (a) Maximize or Minimize f (x, y) = 4x + 6y subject to g(x, y) = x^2 + y^2 − 13 = 0. Let λ be such that ∇f (x, y) = λ∇g(x, y) ⇒ (4, 6) = λ(2x, 2 y) ⇒ λx = 2, λy = 3 ⇒ 2 y = 3x, and g(x, 32 x) = 13x^2 − 13 × 4 = 0 implies x = 2, y = 3, or x = − 2 , y = − 3. Since (2, 3) and (− 2 , −3) are the only two solutions of the above equations: f (2, 3) = 26 and f (− 2 , −3) = −26 gives the absolute minimum and maximum val- ues of f.
(b) Maximize or Minimize f (x 1 , x 2 ,... , xn) = x 1 + x 2 + · · · + xn subject to g(x 1 , x 2 ,... , xn) = x^21 + x^22 + · · · + x^2 n − 1 = 0. Let λ be such that ∇f (x 1 , x 2 ,... , xn) = λ∇g(x 1 , x 2 ,... , xn). (1, 1 ,... , 1) = λ(2x 1 , 2 x 2 ,... , 2 xn) ⇒ x 1 = x 2 = · · · = xn, and g(x 1 , x 1 ,... , x 1 ) = nx^2 i − 1 = 0 gives xi = √^1 n for all i = 1, 2 ,... , n or xi = − √^1 n for all i = 1, 2 ,... , n. The absolute maximum is attained at ( √^1 n , √^1 n ,... , √^1 n ) and the absolute minimum is attained at (− √^1 n , − √^1 n ,... , − √^1 n ).
(c) Maximize or Minimize f (x, y) = e−xy^ subject to x^2 + 4y^2 ≤ 1. The conditions for critical points inside the ellipse x^2 + 4y^2 = 1 are given by: fx(x, y) = 0 and fy (x, y) = 0 which gives −ye−xy^ = 0 and −xe−xy^ = 0, ⇒ x = 0 and y = 0, which lies inside the ellipse. D(0, 0) = fxx(0, 0)fyy(0, 0) − fxy(0, 0)^2 = −e^0 = − 1 < 0, hence (0, 0) is a saddle point of f. For points on the boundary of the ellipse x^2 + 4y^2 = 1, we use Lagrange’s method. Let λ be such that ∇f (x, y) = λ∇g(x, y), where g(x, y) = x^2 + 4y^2 − 1. ⇒ (−ye−xy, −xe−xy^ ) = λ(2x, 8 y) (1) If x = 0 then y = 0, however (0, 0) does not satisfy x^2 + 4y^2 = 1 (also (0, 0) is a saddle point as noted earlier). If x 6 = 0 then y 6 = 0 and we get 2 yx = 8 xy or x^2 = 4y^2 , which gives x = ± √^12 and y = ± 2 √^12. Hence the four possible solutions of (1) are (x, y) = (± √^12 , ± 2 √^12 ).
(x , y ) 1 1
mx +bi
di
(x , y )i i
y
x
Since f ( √^12 , 2 √^12 ) = f (− √^12 , − 2 √^12 ) = e−^ (^14) < f (− √^12 , 2 √^12 ) = f ( √^12 , − 2 √^12 ) = e (^14) . ( √^12 , 2 √^12 ) and (− √^12 , − 2 √^12 ) are absolute minima of f and (− √^12 , 2 √^12 ) and ( √^12 , − 2 √^12 ) are absolute maxima of f.
- Suppose that a scientist has reason to believe that two quantities x and y are related linearly, that is, y = mx + b, at least approximately for some values of m and b. The scientist performs an experiment and collects data in the form of points (x 1 , y 1 ), (x 2 , y 2 ),... , (xn, yn), and then plots these points. The points don’t exactly lie on a straight line, so the scientist wants to find constants m and b such that the line y = mx + b “fits” the points as well as possible. Let di = yi − (mxi + b) be the vertical deviation of the point (xi, yi) from the line. The method of least squares determines m and b so as to minimize
∑^ n
i=
d^2 i , the sum of the squares of these
deviations. Show that according to this method, the line of best fit is obtained when
m
∑^ n
i=
xi + bn =
∑^ n
i=
yi
m
∑^ n
i=
x^2 i + b
∑^ n
i=
xi =
∑^ n
i=
xiyi.
Solution: To find m and b such that f (m, b) =
∑^ n i=
d^2 i is minimum, we have to find
the critical points of f fm(m, b) = 0 gives
∑n i=1 di(−xi) = 0^ (1) and fb(m, b) = 0 gives
∑n i=1 di^ = 0^ (2) From (2) we get m
∑^ n i=
xi + bn =
∑^ n i=
yi.
From (1) we get m
∑^ n
i=
x^2 i + b
∑^ n
i=
xi =
∑^ n
i=
xiyi.
(c) For a given y the area of a cross section of the solid obtained is given by A(y) = π(
y + 1)^2 − π(y + 1)^2. Hence the volume of the solid is given by V =
∫ (^1) 0 A(y)dy^ =^
π
Find the volume of the wedge that is cut from a circular cylinder with unit radius and unit height by a plane that passes through a diameter of the base of the cylinder and through a point on the circumference of its top.
Solution: We show the cylinder and the wedge in the figure above.
To form such a wedge, you may fill a cylindrical glass with water and then drink slowly, tipping the bottom up as you drink, until the half of the bottom of the glass is exposed; the remaining water forms the edge.
Without loss of generality let us assume the plane to cut the base of the cylinder along the x axis, with the centre of the cylinder at the origin (see the second figure in the sequence). Then for a fixed x the cross section of the wedge is an isosceles triangle, since the plane intersects the base of the cylinder at a fixed angle, and at x = 0, the height and the base (given by y) are both equal to 1 (see the third figure in the sequence).
The area of each of these triangles is given by: A(x) = 12 y × y = 12 (1 − x^2 ). Hence the required volume is V =
∫ (^1) − 1 A(x)dx^ =^
∫ (^1) − 1
1 2 (1^ −^ x
(^2) )dx = 2
Prove that the length of one arch of the sine curve y = sin x is equal to half the circumference of the ellipse 2x^2 + y^2 = 2.
Solution: The length of one arch of the sine curve is given by:
L 1 =
∫ (^) π
0
√√ √√ 1 +
( dy dx
) 2 dx
=
∫ (^) π
0
√ 1 + (cos x)^2 dx.
-1 -3 -2 -1 (^0) x 1 2 3
-0.
-0.
-0.
-0.
0
1
y
y=sin(x)
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 x
-1.
-0.
0
1
2
y (-1, 0) (1, 0)
(0, 2)
2x^2 +y^2 =
The length of half the circumference of the ellipse is given by:
L 2 =
∫ (^) +
− 1
√√ √√ 1 +
( 2 x √ 2 − 2 x^2
) 2 dx
∫ (^) +
− 1
√√ √√ 1 +
( 2 x √ 2 − 2 x^2
) 2 dx
∫ (^) +
− 1
√ 1 + x^2 1 − x^2
dx. Take x = cos t then the value of the above integral is equal to: ∫ (^0)
π
√√ √√ 1 + (cos t)^2 (sin t)^2
(− sin t)dt
∫ (^) π
0
√ 1 + (cos t)^2 dt, hence L 1 = L 2.
Aliter: S = 2
∫ (^1)
0
2 πxds = 4π
∫ (^1)
0
(1 − y
(^23) )
(^32)
√√ √√ 1 +
( dx dy
) 2 dy
= 4π
∫ (^1)
0
(1 − y 32 )
(^32) y−^
(^13) dy
= 4π
∫ (^0) π 2 (sin^ θ)
(^3) (−3 cos θ sin θ)dθ (by taking y 23 = cos (^2) θ).
= 12π
∫ π 2
0
(sin θ)^4 (cos θ)dθ =
12 π 5
