How do you use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the given plane x + 8y + 5z = 24?
1 Answer
Explanation:
The volume of a rectangular box is given by the formula
We want to maximize
- Use
#gradf=lambdagradg# , where#f(x,y,z)=xyz# and#g(x,y,z)=x+8y+5z=24#
#gradf=< f_x,f_y,f_z > => < yz, xz, xy >#
#gradg= < g_x,g_y,g_z > => < 1, 8, 5 ># This gives
#< yz, xz, xy > = lambda< 1, 8, 5 >#
Now we set the respective components equal:
#yz=lambda# #xz=8lambda# #xy=5lambda#
There are no general rules for solving systems of equations, so some ingenuity may be required at times. Common strategies involve solving for the variables
#lambda=yz#
#lambda=(xz)/8#
#lambda=(xy)/5#
We can set the equations equal:
#yz=(xz)/8=(xy)/5#
From equations 1 and 2:
#yz=(xz)/8#
#=>8y=x#
From equations 1 and 3:
#yz=(xy)/5#
#=>5z=x#
We can substitute these (convenient) values back into the constraint:
#x+x+x=24#
#=>3x=24#
#=>x=8#
This gives that
You may check these values:
#8+8(1)+5(8/5)=24#
#8+8+8=24#
#24=24#
Therefore, the volume is
A formal explanation (1):
Method of Lagrange Multipliers
To find the maximum and minimum values of
#f(x,y,z)# subject to the constraint#g(x,y,z)=k# [assuming that these extreme values exist and#gradg!=0# on the surface#g(x,y,z)=k# ]:(a) Find all values of
#x,y,z# and#lambda# such that
#gradf(x,y,z)=lambda gradg(x,y,z)#
#g(x,y,z)=k# and
(b) Evaluate
#f# at all points#(x,y,z)# that result from step (a). The largest of these values is the maximum value of#f# ; the smallest is the minimum value of#f# .
- Stewart, James. "14.8 Lagrange Multipliers ." In Calculus: Early Transcendentals, 972. 8th ed. Vol. 1. Boston, MA: Cengage Learning, 2015.
