lagrange V be limited by eq. 2

F(x, y, z) = xyz

G(x, y, z) = 6x + 4y + 3z - 24 = 0

By Lagrange Multiplier, gradF = @gradG

yz = 6@ ... (1)

xz = 4@ ... (2)

xy = 3@ ... (3)

6x + 4y + 3z - 24 = 0 ... (4)

From (1) and (3): 2xy = yz

y = 0 (rejected) or 2x = z

So, put into (2): 2x^2 = 4@

@ = x^2/2 ... (5)

Put (5) into (3): xy = 3(x^2/2)

x = 0 (rejected) or y = 3x/2

So, from (4): 6x + 4(3x/2) + 3(2x) - 24 = 0

18x = 24

x = 4/3

So, y = 2, z = 8/3

Therefore, the largest volume of V is xyz = (4/3)(2)(8/3) = 64/9 sq. units