Question

Solve the optimization problem. hint [see example 2.] maximize p = xyz with x + z = 24 and y + z = 24, and x, y, and z ≥ 0. p =

Answer 1

Let's use the method of Lagrange multipliers. We have the Lagrangian

$L(x,y,z,\lambda_1,\lambda_2)=xyz+\lambda_1(x+z-24)+\lambda_2(y+z-24)$

with partial derivatives (set equal to 0)

$L_x=yz+\lambda_1=0$

$L_y=xz+\lambda_2=0$

$L_z=xy+\lambda_1+\lambda_2=0$

$L_{\lambda_1}=x+z-24\implies x+z=24$

$L_{\lambda_2}=y+z-24\implies y+z=24$

From the last two equations, we have

$x+z=y+z\implies x=y$

so that

$L_z=x^2+\lambda_1+\lambda_2=0$

Then

$L_z-L_y-L_x=x^2-2xz=x(x-2z)=0\implies x=0\text{ or }x=2z$

If $x=0$, then $y=0$ and $z=24$, so one critical point occurs at (0, 0, 24) with a value of $p(0,0,24)=0$.

If $x=2z$, then [te]y=2z[/tex] and $z=8$, making $x=y=16$ and giving another critical point at (16, 16, 8) with a value of $p(16,16,8)=2048$.