Problems based on Lagrange's method of multipliers

1.4.b. Problems based on Lagrange's method of multipliers

Example 1.4b(1): Solve x (y-z) p + y (z - x) q=z (x − y).

Example 1.4b(2): Solve (mz - ny) p + (nx − lz) q = ly - mx.

Solution: Given: (mz - ny) p + (nx - lz) q = ly - mx

Example 1.4b(3) : Sove (3z-4y) p + (4x-2z) q 2y-3x.

Example 1.4b(4) Find the general solution of

Example 1.4b(5) Solve (x² - yz) p + (y² − zx) q = (z² − xy).

Solution: Given: (x² - yz) p + (y² - zx) q = (z² − xy)

Note: The author wishes to thank Dr. B. Jothiram, formerly Asst. Prof. and HOD Maths, Govt. College of Engg., Salem, for having drawn his attention to this elegant method of getting one of the independent solutions.

Example 1.4b(6): Solve (x² + y²+ yz) p + (x² + y² − xz) q = z(x + y).

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Solution: Given: (x² + y²+ yz) p + (x² + y² − xz) q = z(x + y).

This equation is of the form P p + Q q = R

Example 1.4b(7): Solve (y² + z²) p- xyq+xz = 0

Solution: Given: (y²+z²) p-xyq+xzx= 0

This equation is of the form Pp+Qq = R

Example 1.4b(8): Solve (y² + z) p- y(x² +z)q = z(x² –^y2)

Solution: This equation is of the form P p + Qq= R

where P = X x (y²+ z), Q = y (x² + z), R = z ( x² - y²).

The Lagrange's subsidiary equations are

Example 1.4b(9): Solve the equation: (x² - y² - z²) p+ 2 xyq = 2 zx

Solution: Given: (x² - y² - z²) p + 2xyq = 2zx

This equation is of the form Pp+Qq=R

The Lagrange's subsidiary equations are

Example 1.4b(10) : Solve (z²- 2yz-y²) p + (xy + zx) q = xy - zx

Solution : Given: (z²- 2yz-y²) p + (xy + zx) q = xy - zx

This equation is of the form Pp+Qq R where

P = x²-2yz-y², Q = xy + zx, R=xy − zx

The Lagrange's subsidiary equations are

Example 1.4b(11): Solve (y + z) p + (z+x) q = x + y.

Solution: Given (y + z) p + (z+x) q = x+y

This equation is of the form Pp+Qq=R

y+z, Q = z+x, R = x + y

The Lagrange's subsidiary equations are

Example 1.4b(12): Solve (y-xz) p + (yz - x) q = (x + y) (x − y)

Sol. Given: (y-xz) p + (yz − x) q = (x + y) (x − y)

This equation is of the form Pp+Qq=R

where P=(y-xz), Q = yz -x, R= (x + y) (x − y)

Example 1.4b(13): Solve (y-z) p − (2x + y) q = 2x + z.

Solution: Given: (y-z)p (2x+y) q = 2x + z

This equation is of the form Pp+Qq = R

where Py-z, Q = − (2x + y), R = 2x+z

The Lagrange's subsidiary equations are,

Example 1.4b(14): Show that the integral surface of the equation 2y (z-3) p + (2x-z) q = y (2x-3) that passes through the circle x²+ y² = 2x, z=0 is x²+y-²-2x+4z = 0.

Solution :

Given: 2y (z - 3) p + (2x-z) q = y (2x − 3)

This equation is of the form Pp + Qq = R

where P = 2y (z 3), Q = 2xz, R = y (2x-3)

The Lagrange's subsidiary equations are

substitute for a and b from (2) and (3) in (9), we get

x² + y²-z² - 2x + 4z = 0 which is the equation of the required integral surface.