How do you find the maximum value of the function #f(x,y,z)= x+2y-3z# subject to the constraint #z=4x^2+y^2#?

2 Answers
May 2, 2017

Answer:

#17/48#

Explanation:

Forming the lagrangian

#L(x,y,z,lambda)=f(x,y,z)+lambda g(x,y,z)#

with

#g(x,y,z)=4x^2+y^2-z=0#

The stationary points are computed solving

#grad L(x,y,z,lambda)=0#

or

#{(1+8lambda z=0),(2 + 2 lambda y=0),(3 + lambda=0),(4 x^2 + y^2 - z=0):}#

and solving for #(x,y,z,lambda)# we obtain

#{(x=1/24),(y=1/3),(z=17/144),(lambda=-3):}#

and the maximum value is

#17/48#

NOTE: To qualify the stationary point it is necessary to form

#(f @ g) (x,y)= x+2y -3(4x^2+y^2)#

and then calculate

#H=grad^2(f @ g) =((-24, 0),(0, -6))#

As we can observe, #H# has negative eigenvalues indicating that the found solution represents a maximum.

May 2, 2017

Answer:

Explanation:

Given: #f(x,y,z)= x+2y-3z# and the constraint function #g(x,y,z) = 4x^2+y^2-z= 0#

The Lagrange function is:

#L(x,y,z,lambda) = f(x,y,z) + lambdag(x,y,z)#

#L(x,y,z,lambda) = x+2y-3z + 4lambdax^2+lambday^2- lambdaz#

We compute the partial derivatives:

#(delL(x,y,z,lambda))/(delx) = 1+8lambdax#

#(delL(x,y,z,lambda))/(dely) = 2+ 2lambday#

#(delL(x,y,z,lambda))/(delz) =-3-lambda#

#(delL(x,y,z,lambda))/(dellambda) =4x^2+y^2-z#

Set these 4 derivatives equal to zero and then solve them as a system of equation:

#0 = 1+8lambdax" [1]"#

#0 = 2+ 2lambday" [2]"#

#0 =-3-lambda" [3]"#

#0 =4x^2+y^2-z" [4]"#

Solve equation [3] for #lambda#:

#lambda = -3#

Substitute -3 for #lambda# into equation [1]:

#0 = 1+8(-3)x" [1]"#

#x = 1/24#

Substitute -3 for #lambda# into equation [2]:

#0 = 2+ 2(-3)y#

#y = 1/3#

Use equation [4] to find the value of z:

#z = 4(1/24)^2+ (1/3)^2#

#z = 17/144#

#f(1/24,1/3,17/144) = 1/24 + 2/3- 51/144#

#f(1/24,1/3,17/144) = 17/48#

Note: One cannot use the second derivative to test whether the Lagrange multiplier has given you a maximum or a minimum; the only way to determine whether the value is a local maximum is perturbation of values. I will leave that exercise to you.